25.8 - Solution of Homogeneous Differential Equations

In the previous section, we saw homogeneous functions and their degrees. We also saw homogeneous differential equation. In this section, we will see the method to solve such differential equations.

The method can be written in 5 steps:

1. We want to solve the homogeneous differential equation

$\boldsymbol{\frac{dy}{dx}~=~F(x,y)}$

• Since it is a homogeneous differential equation, we can write: $\small{\frac{dy}{dx}~=~F(x,y)~=~x^0\left[g\left(\frac{y}{x} \right) \right]~=~g\left(\frac{y}{x} \right)}$

• In short, we can write: $\small{\frac{dy}{dx}~=~g\left(\frac{y}{x} \right)}$

• Note that, if we are given the homogeneous differential equation $\boldsymbol{\frac{dx}{dy}~=~F(x,y)}$, we must write:
$\small{\frac{dx}{dy}~=~h\left(\frac{x}{y} \right)}$

2. Next, we make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

3. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

4. Substituting (2) and (3) in (1), we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{g(v)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{g(v)~-~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{ \frac{dv}{g(v)~-~v}}    & {~=~}    &{\frac{dx}{x}}    \\
\end{array}}$

5. So we have separated the variables. Integrating both sides, we will get the general solution. In that general solution, we must substitute $\small{\frac{y}{x}~~\text{in the place of}~~v}$

Let us see some solved examples:

Solved example 25.45
Show that the differential equation $\boldsymbol{y'~=~\frac{x+y}{x}}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+y}{x}}$

2. Let $\small{F(x,y)~=~\frac{x+y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x+y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{x\left[\frac{1 ~+~ \frac{y}{x}}{x} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[\frac{x}{x} \right]\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}        &{x^0\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    6    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~1 ~+~ \frac{y}{x}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{x+(vx)}{x}~=~1+v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1+v~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[1 \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[1 \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{v~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{v}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{v}    & {~=~}&{\log \left|x \right|~+~\rm{C}_3}    \\
\end{array}}$

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{v}    & {~=~}    &{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\frac{y}{x}}    & {~=~}    &{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{y}    & {~=~}    &{x\log \left|x \right|~+~x \rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{y}    & {~=~}    &{x\log \left|x \right|~+~{\rm{C}}\,x}    \\
\end{array}}$

Solved example 25.46
Show that the differential equation $\boldsymbol{(x-y)dy~-~(x+y)dx~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+y}{x-y}}$

2. Let $\small{F(x,y)~=~\frac{x+y}{x-y}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x+y}{x-y}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x}{x} \right]\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v}{1-v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v}{1-v}~-~v~=~\frac{1+v-v+v^2}{1-v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{1-v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{1-v}{1+v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1-v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\int{\left[\frac{1}{1+v^2} \right]dv}~-~\int{\left[\frac{v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}    &{\tan^{-1}v~-~\frac{1}{2}\log \left|1+v^2 \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\tan^{-1}v~-~\frac{1}{2}\log \left|1+v^2 \right|}    & {~=~}&{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{2 \tan^{-1}v~-~\log \left|1+v^2 \right|}    & {~=~}&{2\log \left|x \right|~+~2\rm{C}_3}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{2 \tan^{-1}v}    & {~=~}&{\log \left|1+v^2 \right|~+~2\log \left|x \right|~+~\rm{C}_4}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{2 \tan^{-1}v}    & {~=~}&{\log \left|\left(1+v^2 \right)x^2 \right|~+~\rm{C}_4}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(2) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{2 \tan^{-1}v}    & {~=~}    &{\log \left|\left(1+v^2 \right)x^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{2 \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|\left(1+\frac{y^2}{x^2} \right)x^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{2 \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|x^2 + y^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{ \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\frac{1}{2}\log \left|x^2 + y^2 \right|~+~\left(\frac{1}{2} \right)\rm{C}_4}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{ \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\frac{1}{2}\log \left|x^2 + y^2 \right|~+~\rm{C}}    \\
\end{array}}$

Solved example 25.47
Show that the differential equation $\boldsymbol{(x-y)\frac{dy}{dx}~=~x+2y}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+2y}{x-y}}$

2. Let $\small{F(x,y)~=~\frac{x+2y}{x-y}}$
• We need to show that, this is a homogeneous function of degree zero.
• We already showed it in solved example 25.43 of the previous section. So the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+2y}{x-y}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{x+2(vx)}{x-vx}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x+2(vx)}{x-vx}~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x~+~2vx~-~vx~+~v^2x}{x-vx}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x~+~vx~+~v^2x}{x-vx}~=~\frac{1~+~v~+~v^2}{1-v}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\frac{1-v}{1~+~v~+~v^2} \right]dv}    & {~=~}    &{\frac{dx}{x}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\left[\frac{v-1}{1~+~v~+~v^2} \right]dv}    & {~=~}    &{\left[\frac{-1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{v-1}{1~+~v~+~v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\int{\left[\frac{2v + 1 - 3}{2(1~+~v~+~v^2)} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\frac{1}{2} \int{\left[\frac{2v + 1}{1~+~v~+~v^2} \right]dv}~-~\frac{3}{2} \int{\left[\frac{1}{1~+~v~+~v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left| v^2 + v+ 1  \right|~-~\left(\frac{3}{2} \right)\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_1}    & {~=~}&{-\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left|v^2 + v+ 1 \right|~+~\log \left|x \right|}    & {~=~}&{\left(\frac{3}{2} \right)\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left|v^2 + v+ 1  \right|~+~\log \left|x \right|}    & {~=~}&{\sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(3) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{1}{2} \log \left|v^2 + v+ 1  \right|~+~\log \left|x \right|}    & {~=~}    &{\sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|v^2 + v+ 1  \right|~+~2 \log \left|x \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|\frac{y^2}{x^2} + \frac{y}{x}+ 1 \right|~+~\log \left|x^2 \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\left(\frac{y^2}{x^2} + \frac{y}{x}+ 1 \right)x^2  \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\log \left|y^2 + xy + x^2 \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~ \rm{C}}    \\
\end{array}}$

Solved example 25.48
Show that the differential equation $\boldsymbol{(x^2 + xy)dy~=~(x^2 + y^2)dx}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x^2+y^2}{x^2+xy}}$

2. Let $\small{F(x,y)~=~\frac{x^2+y^2}{x^2+xy}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x^2+y^2}{x^2+xy}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{1+v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2~}{1+v}~-~v~=~\frac{1+v^2-v-v^2}{1+v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v}{1+v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{1+v}{1-v} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1+v}{1-v} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\int{\left[\frac{1}{1-v} \right]dv}~+~\int{\left[\frac{v}{1-v} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}    &{(-1)\log \left|1-v \right|~+~\left[-v~-~\log \left|1-v \right| \right]~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{-\log \left|1-v \right|~-~v~-~\log \left|1-v \right| ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{-2\log \left|1-v \right|~-~v ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{\log \left|x \right|~+~2\log \left|1-v \right|~+~v}    & {~=~}&{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{\log \left|x(1-v)^2 \right|~+~v}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(2) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|x(1-v)^2 \right|~+~v}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|x\left[1-\left(\frac{y}{x} \right) \right]^2 \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|x\left[\frac{x-y}{x} \right]^2 \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|}    & {~=~}    &{\log \left|\rm{C}\right|~-~\frac{y}{x}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|~-~\log \left|\rm{C}\right|}    & {~=~}    &{~-~\frac{y}{x}}    \\
{~\color{magenta}    7    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{{\rm{C}}x} \right|}    & {~=~}    &{~-~\frac{y}{x}}    \\
{~\color{magenta}    8    }&{{\Rightarrow}}&{\frac{(x-y)^2}{{\rm{C}}x}}    & {~=~}    &{e^{-\frac{y}{x}}}    \\
{~\color{magenta}    9    }&{{\Rightarrow}}&{(x-y)^2}    & {~=~}    &{{\rm{C}}x\,e^{-\frac{y}{x}}}    \\
\end{array}}$

Solved example 25.49
Show that the differential equation $\boldsymbol{(x^2 - y^2)dx~+~2xy\,dy~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x^2 - y^2)dx~+~2xy\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{2xy\,dy}    & {~=~}    &{(-1)(x^2 - y^2)dx}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{(-1)(x^2 - y^2)}{2xy}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~\frac{(-1)(x^2 - y^2)}{2xy}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{(-1)(x^2 - y^2)}{2xy}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~-~ \frac{y^2}{x^2}}{(-2)\frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~-~ \frac{y^2}{x^2}}{(-2) \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v^2}{(-2)v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v^2~}{(-2)v}~-~v~=~\frac{1-v^2+2v^2}{(-2)v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{(-2)v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{(-2)v}{1+v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{(-2)v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{(-2)\left(\frac{1}{2} \right)\log \left|1+v^2 \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{-\log \left|1+v^2 \right| ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\log \left|x \right|~+~\log \left|1+v^2 \right|}    & {~=~}&{\rm{C}_1~-~\rm{C}_2~=~\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|x \right|~+~\log \left|1+v^2 \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|x\left[1+\left(\frac{y}{x} \right)^2 \right] \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|x\left[\frac{x^2 + y^2}{x^2} \right] \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{x^2 + y^2}{x} \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\frac{x^2 + y^2}{x}}    & {~=~}    &{e^{\rm{C}_3}~=~\rm{C}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{x^2~+~y^2}    & {~=~}    &{{\rm{C}}x}    \\
\end{array}}$

Solved example 25.50
Show that the differential equation $\boldsymbol{x^2\,\frac{dy}{dx}~=~x^2~-~2y^2~+~xy}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2\,\frac{dy}{dx}}    & {~=~}    &{x^2~-~2y^2~+~xy}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{1~-~ 2\left(v \right)^2~+~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1 - 2 v^2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left[\frac{1}{1 - 2v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1}{1 - 2v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2(y/x) + \sqrt 2}{2(y/x) - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{\sqrt 2(y/x) + 1}{\sqrt 2(y/x) - 1}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{\sqrt 2 \,y ~+~ x}{\sqrt 2 \,y ~-~ x}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

Solved example 25.51
Show that the differential equation $\boldsymbol{x dy ~-~y dx~=~\sqrt{x^2 + y^2}\,dx}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x dy ~-~y dx}    & {~=~}    &{\sqrt{x^2 + y^2}\,dx}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{x dy }    & {~=~}    &{\sqrt{x^2 + y^2}\,dx~+~y dx}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{x dy }    & {~=~}    &{\left[\sqrt{x^2 + y^2}~+~y \right]dx}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~\frac{\sqrt{x^2 + y^2}~+~y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(\lambda x,\lambda y)}    & {~=~}    &{\frac{\sqrt{(\lambda x)^2 + (\lambda y)^2}~+~\lambda y}{\lambda x}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{\lambda \sqrt{x^2 + y^2}~+~\lambda y}{\lambda x}}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\lambda^0\left[F(x,y) \right]}    \\
\end{array}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n\left[F(x,y) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + v^2\,x^2}~+~vx}{x}~=~\sqrt{1+v^2}~+~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\sqrt{1 + v^2}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left[\frac{1}{\sqrt{1 + v^2}} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1}{\sqrt{1 + v^2}} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\log \left|\sqrt{1 + v^2}~+~v \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\log \left|\sqrt{1 + v^2}~+~v \right|~-~\log \left|x \right|}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|\sqrt{1 + v^2}~+~v \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|\sqrt{1 + (y/x)^2}~+~(y/x) \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|\frac{\sqrt{x^2 + y^2}~+~y}{x} \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{\sqrt{x^2 + y^2}~+~y}{x^2} \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\frac{\sqrt{x^2 + y^2}~+~y}{x^2}}    & {~=~}    &{e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{\frac{\sqrt{x^2 + y^2}~+~y}{x^2}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    7    }&{{\Rightarrow}}&{\sqrt{x^2 + y^2}~+~y}    & {~=~}    &{{\rm{C}}x^2}    \\
\end{array}}$

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

23.21 - Definite Integrals

In the previous section, we completed a discussion on three standard integrals. In this section, we will see definite integrals.

Let us first see a method to calculate the area of a region. The region that we are considering here, is a special kind of region because, one of it's sides is not a straight line. The method can be written in 15 steps:

1. In fig.23.2 below, the red curve is the graph of y = f(x).

Fig.23.2

• We assume that, for all input x values, f(x) is non negative. So the graph is above the x-axis.
• There are two vertical lines: x = a and x = b. They are drawn in green color.
• So we have a region bounded by the four items below:
   ♦ The curve y = f(x)
   ♦ The vertical line x = a
   ♦ The vertical line x = b
   ♦ The horizontal line y = 0 (x-axis)
• Our aim is to find the area of this region.

2. In fig.23.3 below, the region is divided into strips.

Fig.23.3

• The strips have the following 4 properties:
(i) All strips are formed by vertical lines.
(ii) All strips have the same width. We will denote this width as 'h'.
(iii) There is a total of n strips.
(iv) Each of the n strips, has it's own position.
   ♦ The left edge of the first strip is at x = x₀ = a
   ♦ The left edge of the second strip is at x = x₁
   ♦ The left edge of the third strip is at x = x₂
   ♦ The left edge of the fourth strip is at x = x₃
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the r^th strip is at x = x_r−1
   ♦ The left edge of the (r+1)^th strip is at x = x_r
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the n^th strip is at x = x_n−1
   ♦ The right edge of the n^th strip is at x = x_n = b

3. Based on the above properties of the strips, we can write:
   ♦ x₀ = a
   ♦ x₁ = a+h
   ♦ x₂ = a+h+h = a+2h
   ♦ x₃ = a+2h+h = a+3h
   ♦ x₄ = a+3h+h = a+4h
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_r = a+rh
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_n = a+nh
4. Now we can write an expression for n.
• We have $\small{x_n~=~a+nh~=~b}$
$\small{\Rightarrow nh~=~b-a}$
$\small{\Rightarrow n~=~\frac{b-a}{h}}$

5. From the above expression, we get an important information:
• h is in the denominator. So when n increases, h decreases.
• Consequently, when n is close to infinity, h will be close to zero.

6. Now select any convenient strip as the r^th strip. In fig.23.3, it is named as ABCD.
• Note that, none of the strips is a rectangle. This is because, the top edge of any selected strip is not horizontal. Our present strip ABCD is also not a rectangle because, the top edge CD is not horizontal.

7. Since CD is not horizontal, we draw two new horizontal lines:
   ♦ Horizontal line CL through C
   ♦ Horizontal line DM through D
• Now we have three regions:
   ♦ Rectangle ABDM
   ♦ Region ABDCA
   ♦ Rectangle ABLC

8. Let us compare the areas of the three regions. From the fig., it is clear that:
• Area of region ABDCA is less than the area of the rectangle ABDM
• Area of region ABDCA is greater than the area of the rectangle ABLC

9. Consider the situation when n is close to infinity.
• In such a situation,
   ♦ h will be very close to zero.
   ♦ B will be very close to A
   ♦ L will be very close to C
   ♦ Segment MD will be very close to segment CL
• In such a situation, all three areas mentioned in (8), can be considered to be equal.

10. It follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABLC.
• Area of rectangle ABLC = AC  × AB = f(x_r-1)  × h
• Here we are using the lower rectangle ABLC to get the area of the region ABCDA. Length of the rectangle is left edge AC.

11. Similarly, it follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABDM.
• Area of rectangle ABDM = BD  × AB = f(x_r)  × h
• Here we are using the upper rectangle ABDM to get the area of the region ABCDA. Length of the rectangle is right edge BD.

12. The information in both (10) and (11) are applicable to all strips.
• Let us first apply the information in (10).
   ♦ Area of first strip = left edge × base = f(x₀) × h
   ♦ Area of second strip = left edge × base = f(x₁) × h
   ♦ Area of third strip = left edge × base = f(x₂) × h
   ♦ Area of fourth strip = left edge × base = f(x₃) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = left edge × base = f(x_r−1) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = left edge × base = f(x_n−1) × h
• So total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]= $\small{h\sum\limits_{r=0}^{r=n-1}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
Total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]
= h[f(a) + f(a+h) + f(a+2h) + . . . +f(a+(n−1)h)]
= $\small{h\sum\limits_{k=0}^{k=n-1}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a)~+~f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+(n-1)h) \right]}$

13. Let us now apply the information in (11).
   ♦ Area of first strip = right edge × base = f(x₁) × h
   ♦ Area of second strip = right edge × base = f(x₂) × h
   ♦ Area of third strip = right edge × base = f(x₃) × h
   ♦ Area of fourth strip = right edge × base = f(x₄) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = right edge × base = f(x_r) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = right edge × base = f(x_n) × h
• So total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)] = $\small{h\sum\limits_{r=1}^{r=n}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
• Total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)]
= h[f(a+h) + f(a+2h) + f(a+3h) + . . . +f(a+nh)]
= $\small{h\sum\limits_{k=1}^{k=n}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

14. Let us compare the methods.
• Step (12) gives us a method to find the area of PRSQP.
It uses the left edge of each rectangle.
• Step (13) also gives us a method to find the area of PRSQP.
It uses the right edge of each rectangle.
• We can use any one of the two methods. It is convenient to use the method in (13). This is because, it has f(a+nh) instead of f(a+(n−1)h)

15. So we can write:
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

• Using the result in step (4), we can replace the 'h' outside the square brackets. We get:
Area of PRSQP =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

---

Let us see a solved example:
Solved Example 23.69
Find the area bounded by the four items:
   ♦ The curve y = f(x) = x².
   ♦ The vertical line x = 2
   ♦ The vertical line x = 3
   ♦ The horizontal line y = 0 (x-axis)
Solution:
1. A rough sketch is shown in the fig.23.4 below:

Fig.23.4

• Our task is to find the area of the violet shaded area.
2. We have the formula for area:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$
• In our present case, a = 2 and b = 3
So $\small{h~=~\frac{b-a}{n}~=~\frac{3-2}{n}~=~\frac{1}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$

= $\small{\lim_{n\rightarrow \infty} \frac{3-2}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$

= $\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[(a+\frac{1}{n})^2~+~(a+\frac{2}{n})^2~+~~.~.~.~+(a+\frac{n}{n})^2 \right]}$

4. Let us determine the quantity inside the square brackets:

$\small{ \left[\left(a^2 + \frac{2a}{n} + \frac{1^2}{n^2} \right)~+~\left(a^2 + \frac{4a}{n} + \frac{2^2}{n^2} \right)~+~.~.~.~+~\left(a^2 + \frac{2an}{n} + \frac{n^2}{n^2} \right) \right]}$

5. So we have to do three summations:

(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(2^2)~=~4n}}$

$\small{~~~~~=~\small{n a^2~=~n(2^2)~=~4n}}$

(ii) $\small{\frac{2a}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{2a}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{a(n+1)}{2}~=~\frac{2(n+1)}{2}~=~n+1}$

(Here we use the technique of arithmetic progression)

(iii) $\small{\frac{1}{n^2}\left(1^2~+~2^2~+~3^2~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{1}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

(Here we use the technique that we saw in section 9.5)

6. So the total of three summations is:

$\small{4n~+~n+1~+~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

$\small{~=~\frac{2}{n}~+~\frac{19\,n}{3}-\frac{9}{2}+\frac{1}{6n}~=~\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2}}$

7. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2} \right]}$

$\small{~=~\lim_{n\rightarrow \infty}  \left[\frac{13}{6n^2}~+~\frac{19}{3}-\frac{9}{2n} \right]~=~\frac{19}{3}}$

---

Let us write the definition of definite integral. It can be written in 3 steps:
1. We have seen a method to find the area.
• Symbolically, this area is denoted as $\small{\int_a^b{\left[f(x) \right]dx}}$.
   ♦ This is the definite integral of f(x).
2. So we can write:
Area = $\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right] ~=~\int_a^b{\left[f(x) \right]dx}}$
3. Within the square brackets, we have a sum. We are taking the limiting value of that sum. So it is a limit of sum.
• Then we can define the definite integral $\small{\int_a^b{\left[f(x) \right]dx}}$, as the limit of sum.

---

Now we will write an important point. It can be written in 3 steps:
1. In the above solved example, we calculated the definite integral of f(x) = x²
2. It is clear that the variable associated with the function f is x.
• But we applied the limit to the variable n.
3. We see that, the result will depend upon f(x), a and b. The result will not depend upon n.
• So n is called the variable of integration. It is also called the dummy variable.

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

23.19 - Derivation of Some Standard Integrals

In the previous section, we completed a discussion on the method of integration by parts. In this section, we will see three standard integrals.

I. The integral $\bf{\int{\left[\sqrt{x^2 - a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 - a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 - a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2-a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2-a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2 + a^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2}{\sqrt{x^2-a^2}}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 - a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 - a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\sqrt{x^2-a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2-a^2} \big]dx}~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2-a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula IV that we saw in section 23.8. Thus we get:

$\small{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

II. The integral $\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 + a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 + a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2+a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2+a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2 - a^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2}{\sqrt{x^2-a^2}}~-~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2}~-~\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 + a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 + a^2} ~-~\int{\big[\sqrt{x^2+a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2+a^2} \big]dx}~=~x \sqrt{x^2 + a^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2+a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula VI that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

III. The integral $\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{a^2 - x^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{a^2 - x^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-x}{\sqrt{a^2-x^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-x}{\sqrt{a^2-x^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2 - a^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2}{\sqrt{a^2-x^2}}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x\sqrt{a^2 - x^2} \big]~-~\bigg[\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} \bigg]}$

$\small{~=~x\sqrt{a^2 - x^2} ~-~\int{\big[\sqrt{a^2-x^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{a^2-x^2} \big]dx}~=~x\sqrt{a^2 - x^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{a^2-x^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula V that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

---

Now we will see some solved examples.

Solved example 23.61
Find $\small{\int{\left[\sqrt{x^2 +2x+ 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +2x+ 1+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 2^2} \right]dx}}$

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ 4} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = 2

• So we get:

$\small{\int{\left[\sqrt{u^2 + 2^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + 2^2} ~+~\frac{2^2}{2} \log \left|u+\sqrt{u^2 + 2^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+1)^2 + 2^2} \right]dx}~=~\frac{x+1}{2}\sqrt{(x+1)^2 + 2^2} ~+~\frac{2^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2 + 2^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 2x + 5} \right]dx}~=~\frac{x+1}{2}\sqrt{x^2 + 2x + 5} ~+~2 \log \left|(x+1)+\sqrt{x^2 + 2x + 5} \right|~+~\rm{C}}$

Solved Example 23.62
Find $\small{\int{\left[\sqrt{x^2 +4x+ 6} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4+ 2} \right]dx}~=~\int{\left[\sqrt{(x+2)^2+ 2} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ (\sqrt{2})^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ (\sqrt{2})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = √2

• So we get:

$\small{\int{\left[\sqrt{u^2 + (\sqrt{2})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|u+\sqrt{u^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 + (\sqrt{2})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+1)^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|(x+2)+\sqrt{(x+2)^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 6} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 6} ~+~ \log \left|x+2+\sqrt{x^2 + 4x + 6} \right|~+~\rm{C}}$

---

We have seen the three standard integrals. In the next section, we will see a few more solved examples.

Previous

Contents

Next

Copyright©2025 Higher secondary mathematics.blogspot.com

23.18 - Integration by Parts - Special case

In the previous section, we saw the method of Integration by parts. We saw some solved examples also. In this section, we will see a special case of this method.

Integral of the type $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

This can be calculated in 3 steps:

1. We have: $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

$\small{~=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$

• There are two terms. We will denote the first one as I₁.

2. Let us calculate I₁:

(i) Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{f(x)}$

   ♦ Let second function be: g(x) = $\small{e^x}$

(ii) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$

(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[f(x) \, \left(e^x \right) \big]}$

• This is the first term.

(iv) $\small{f'(x)~=~f'(x)}$

(v) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[f'(x)\,\left(e^x \right)  \big]dx}}$

• This is second term.

(vi) So we get:

I₁ = $\small{\text{First term - Second term}}$

$\small{~=~\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right)  \big]dx} \bigg]}$

3. So from (1), we get:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

$\small{=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$

= I₁ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$

= $\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right)  \big]dx} \bigg]$ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$

= $\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

---

Let us see some solved examples:

Solved Example 23.57
Find $\small{\int{\big[e^x \left[\sin x + \cos x \right]\big]dx}}$
Solution:
1. Let f(x) = $\small{\sin x}$

• Then $\small{f'(x)~=~\cos x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sin x~+~\rm{C}}$

Solved Example 23.58
Find $\small{\int{\left[e^x\,\sec x (1 + \tan x) \right]dx}}$
Solution:
1. The given integral can be written as:

$\small{\int{\big[e^x \left[\sec x + \sec x \tan x \right]\big]dx}}$

Let f(x) = $\small{\sec x}$

• Then $\small{f'(x)~=~\sec x \tan x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sec x~+~\rm{C}}$

Solved Example 23.59
Find $\small{\int{\left[\frac{x e^x}{(1+x)^2} \right]dx}}$
Solution:
1. The $\small{\left[\frac{x}{(1+x)^2} \right]}$ portion can be rearranged as shown below:

$\small{\frac{x}{(1+x)^2} = \frac{1+x-1}{(1+x)^2} = \big[\frac{1+x}{(1+x)^2}~+~\frac{(-1)}{(1+x)^2}\big] = \big[\frac{1}{1+x}~+~\frac{(-1)}{(1+x)^2}\big]}$

2. Now, $\small{\frac{(-1)}{(1+x)^2}}$ is the derivative of $\small{\frac{1}{1+x}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

Where $\small{f(x) = \frac{1}{1+x}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{1}{1+x} \right]~+~\rm{C}}$

Solved example 23.60
Find $\small{(i)~\int{\bigg[e^x \left[\tan^{-1}x \,+\,\frac{1}{1+x^2} \right]\bigg]dx}~~~~~~(ii)~\int{\left[\frac{(x^2 + 1)e^x}{(x+1)^2} \right]dx}}$
Solution:
Part (i):
1. Let f(x) = $\small{\tan^{-1} x}$
• Then $\small{f'(x)~=~\frac{1}{1 + x^2}}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \tan^{-1} x~+~\rm{C}}$

Part (ii):

1. The $\small{\left[\frac{(x^2 + 1)}{(x+1)^2} \right]}$ portion can be rearranged as shown below:

$\small{\frac{x^2 + 1}{(x+1)^2} = \frac{x^2 + 1-1+1}{(x+1)^2} = \frac{x^2 -1+2}{(x+1)^2} = \frac{x^2 -1}{(x+1)^2} + \frac{2}{(x+1)^2}}$

$\small{= \frac{(x+1)(x-1)}{(x+1)^2} + \frac{2}{(x+1)^2}= \frac{x-1}{x+1} + \frac{2}{(x+1)^2}}$

2. Now, $\small{\frac{2}{(x+1)^2}}$ is the derivative of $\small{\frac{x-1}{x+1}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

Where $\small{f(x) = \frac{x-1}{x+1}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{x-1}{x+1} \right]~+~\rm{C}}$

---

The link below gives a few more solved examples:

Exercise 23.6

---

We have completed a discussion on integration by parts. In the next section, we will see some standard integrals.

Previous

Contents

Next

Copyright©2025 Higher secondary mathematics.blogspot.com

23.17 - Solved Examples on Integration by Parts

In the previous section, we saw the basic details about Integration by parts. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.45
Find $\small{\int{\left[x \, \log 2x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log 2x

   ♦ Let second function be: g(x) = x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x\right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log 2x \, \left(\frac{x^2}{2} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{x^2}{4}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \, \log 2x \right]dx}~=~\text{First term - Second term}~=~\log 2x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$

Solved Example 23.46
Find $\small{\int{\left[x^2 \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = $\small{x^2}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2\right]dx}~=~\frac{x^3}{3}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3} \right) \big]}$

• This is the first term.

4. $\small{~f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3} \right)  \big]dx}~=~\frac{x^3}{9}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x^2 \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^3}{3} \right)~-~\frac{x^3}{9}~+~\rm{C}}$

Solved Example 23.47
Find $\small{\int{\left[x \, \sin^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = sin u

$\small{\Rightarrow \frac{dx}{du}\,=\,\cos u \Rightarrow\frac{dx}{\cos u}\,=\,du}$

Also, $\small{~\sqrt{1 - x^2}\,=\,\cos u}$

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\sin^2 u}{\cos u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\sin^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\sin^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 - \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 - \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,-\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin u \cos u}{4}~=~\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4} \big]}$

$\small{~=~\sin^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,+\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$

Solved Example 23.48
Find $\small{\int{\left[x \, \tan^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\tan^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{1+x^2}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{1+x^2}  \big]dx}}$

(i) Let x = tan u

$\small{\Rightarrow \frac{dx}{du}\,=\,\sec^2 u \Rightarrow\frac{dx}{\sec^2 u}\,=\,du}$

Also, $\small{~1 + x^2\,=\,\sec^2 u}$

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\tan^2 u}{\sec^2 u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\tan^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\tan^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\sec^2 u \,-\,1  \big]du}\,=\,\frac{\tan u}{2}\,-\,\frac{u}{2}}$

$\small{\,=\,\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}\big]}$

$\small{~=~\tan^{-1}x \, \left(\frac{x^2}{2}\,+\,\frac{1}{2}\right)\,-\,\frac{x}{2}\,+\,\rm{C}}$

Solved Example 23.49
Find $\small{\int{\left[x \, \cos^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\cos^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{-1}{2}\int{\big[\frac{1}{\sqrt{1-x^2}}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{-1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = cos u

$\small{\Rightarrow \frac{dx}{du}\,=\,-\sin u \Rightarrow\frac{-dx}{\sin u}\,=\,du}$

Also, $\small{~\sqrt{1 - x^2}\,=\,\sin u}$

(ii) So we want:

$\small{\frac{-1}{2}\int{\big[\frac{\cos^2 u}{\sin u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\cos^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\cos^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 + \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 + \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,+\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin u \cos u}{4}~=~\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \cos^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4} \big]}$

$\small{~=~\cos^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,-\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$

Solved Example 23.50
Find $\small{\int{\left[\sin^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \right)  \big]dx}~=~-\sqrt{1-x^2}}$

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\sin^{-1}x \, \left(x\right) \big]~-~\big[-\sqrt{1-x^2} \big]}$

$\small{~=~x \sin^{-1}x~+~\sqrt{1-x^2}\,+\,\rm{C}}$

Solved Example 23.51
Find $\small{\int{\left[(\sin^{-1}x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{\sin^{-1}x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin^{-1}x \right]dx}~=~x \sin^{-1}x~+~\sqrt{1-x^2}}$

(See solved example 23.50 above)

3. $\small{\big[f(x) \left(A \right) \big]~=~\sin^{-1}x \big[x \sin^{-1}x~+~\sqrt{1-x^2} \big]}$

$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\,  \left(\sqrt{1-x^2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \sin^{-1}x~+~\sqrt{1-x^2} \right)  \big]dx}}$

$\small{~=~\int{\big[\frac{x \sin^{-1}x}{\sqrt{1-x^2}}~+~1  \big]dx}~=~x\,-\,(\sqrt{1-x^2}) \sin^{-1}x~+~x}$

$\small{~=~2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x}$

(See solved example 23.38 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[(\sin^{-1} x)^2 \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\,  \left(\sqrt{1-x^2}\right) \big]~-~\big[2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x \big]}$

$\small{~=~x (\sin^{-1}x)^2~+~2 \sin^{-1}x\,  \left(\sqrt{1-x^2}\right) ~-~2x}$

Solved Example 23.52
Find $\small{\int{\left[\frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\cos^{-1}x}$

   ♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\frac{x }{\sqrt{1 - x^2}} \right]dx}~=~-\sqrt{1-x^2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1 - x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1 - x^2}}\,\left(-\sqrt{1-x^2} \right)  \big]dx}~=~x}$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]~-~\big[x \big]~+~\rm{C}}$

$\small{~=~- \big[x~+~(\cos^{-1}x) \,\sqrt{1-x^2}\big]~+~\rm{C}}$

Solved Example 23.53
Find $\small{\int{\left[x \sec^2 x \right]dx}}$
Solution:
$\small{\int{\left[x \sec^2 x \right]dx}}$

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{x}$

   ♦ Let second function be: g(x) = $\small{\sec^2 x }$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sec^2 x \right]dx}~=~\tan x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \tan x \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\tan x \right)  \big]dx}~=~\log \left|\sec x \right| }$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \tan x \big]~-~\big[\log \left|\sec x \right| \big]~+~\rm{C}}$

$\small{~=~x \tan x ~+~\log \left|\frac{1}{\sec x} \right| ~+~\rm{C}}$

$\small{~=~x \tan x ~+~\log \left|\cos x \right| ~+~\rm{C}}$

Solved Example 23.54
Find $\small{\int{\left[\tan^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\tan^{-1} x}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(x \right)  \big]dx}~=~\frac{1}{2}\log \left|1+x^2 \right|}$

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\tan^{-1}x \, \left(x\right) \big]~-~\big[\frac{1}{2}\log \left|1+x^2 \right| \big]}$

$\small{~=~x \tan^{-1}x~-~\frac{1}{2}\log \left|1+x^2 \right|\,+\,\rm{C}}$

Solved Example 23.55
Find $\small{\int{\left[x (\log x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{(\log x)^2}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{2 \log \left|x \right|}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{2 \log \left|x \right|}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}}$

$\small{~=~\int{\big[x\log \left|x \right|  \big]dx}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}}$

(See solved example 23.44 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[x (\log x)^2 \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]~-~\big[\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4} \big]}$

$\small{~=~(\log x)^2 \, \left(\frac{x^2}{2}\right) ~-~\log x \, \left(\frac{x^2}{2} \right)~+~\frac{x^2}{4} }$

Solved Example 23.56
Find $\small{\int{\left[(x^2 + 1) \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = $\small{x^2 + 1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2 + 1\right]dx}~=~\frac{x^3}{3}~+~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3}~+~x \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3}~+~x \right)  \big]dx}~=~\frac{x^3}{9}~+~x}$

• This is the second term.

6. So we get:

$\small{\int{\left[(x^2 + 1) \log x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\log x \, \left(\frac{x^3}{3}~+~x \right)~-~\frac{x^3}{9}~-~x~+~\rm{C}}$

---

We have seen the method of integration by parts. In the next section, we will see a special case in this method.

Previous

Contents

Next

Copyright©2025 Higher secondary mathematics.blogspot.com