23.5 - Solved Examples on Integration by Substitution

In the previous section, we saw some standard integrals of trigonometric functions. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.7
Find the following integrals:
(i) $\small{\int{\left[(4x+2) \sqrt{x^2 + x + 1} \right]dx}}$

(ii) $\small{\int{\left[\frac{1}{1\,-\, \tan x} \right]dx}}$

(iii) $\small{\int{\left[\frac{1}{1\,+\, \cot x} \right]dx}}$

(iv) $\small{\int{\left[\frac{1}{x\,+\, x \log x} \right]dx}}$

Solution:
Part (i):
1. The derivative of (x²+x+1) is 2x+1.
• So we put u = x²+x+1
⇒ $\small{\frac{du}{dx}~=~2x\,+\,1}$
⇒ (2x+1)dx = du

2. So we want:
$\small{\int{\left[(4x+2) \sqrt{x^2 + x + 1} \right]dx}~=~\int{\left[2 \sqrt{u} \right]du}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[2 \sqrt{u}\right] \, du}}    & {~=~}    &{2 \left[\frac{u^{3/2}}{3/2}~+~C_1 \right]~=~\frac{4 u^{3/2}}{3}}~+~C_2    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4(x^2\,+\,x\,+\,1)^{3/2}}{3}~+~C}    \\
\end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,-\,\tan x}}    & {~=~}    &{\frac{1}{1\,-\,(\sin x / \cos x)}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\cos x\,-\,\sin x}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin (\pi/2 \,-\, x)\,-\,\sin x}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{2 \cos(\pi/4) \sin(\pi/4\,-\, x)}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sqrt{2} \sin(\pi/4 \,-\, x)}}    \\
\end{array}}$                           

• Derivative of (π/4 − x) w.r.t x is −1.
• So we put u = (π/4 −x)
⇒ $\small{\frac{du}{dx}~=~-1}$
⇒ −dx = du
• Also, since u = (π/4 −x), we get: x = π/4 − u
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{(-1)(-1)\cos x}{\sqrt{2} \sin(\pi/4 \,-\, x)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-\cos (\pi/4 \,-\, u)}{\sqrt{2} \sin u}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-\cos (\pi/4) \cos u ~-~\sin (\pi/4) \sin u}{\sqrt{2} \sin u}\right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cos u ~-~(1/\sqrt 2)\sin u }{\sqrt{2} \sin u}\right] \, du}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cot u~-~(1/\sqrt 2) }{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2)(\cot u ~+~1)}{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1~+~\cot u)}{2}\right] \, du}}    \\
\end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{-1~-~\cot u}{2}\right] \, du}}    & {~=~}    &{\int{\left[\frac{-1}{2}\right] \, du}~-~\int{\left[\frac{\cot u}{2}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \int{\left[1 \right] \, du}~-~\frac{1}{2} \int{\left[\cot u \right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \left[u\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sin u |\,+\, C_2 \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\frac{1}{2} \left[(\pi/4)\,-\,x\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sin ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,-\,C_1 \right]~+~\frac{1}{2} \left[-\log |\sin ((\pi/4)\,-\,x) |\,-\, C_2 \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log |\sin ((\pi/4)\,-\,x) |\,+\, C_4 \right]}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|\sin (\pi/4) \, \cos x~-~\cos (\pi/4)  \,\sin x \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2) \, \cos x ~-~  (1/\sqrt 2) \,\sin x \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2)(\cos x ~-~\sin x)  \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log (1/\sqrt 2)~-~ \log \left|(\cos x ~-~\sin x)  \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(\cos x ~-~\sin x)  \right |\,+\, C_5 \right]}    \\
{~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{C_3}{2}~-~\frac{\log \left|(\cos x ~-~\sin x)  \right |}{2}\,+\,\frac{C_5}{2}}    \\
{~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,-\,\frac{\log \left|(\cos x ~-~\sin x)  \right |}{2}\,+\,C}    \\
\end{array}}$                            
                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,+\,\cot x}}    & {~=~}    &{\frac{1}{1\,+\,(\cos x / \sin x)}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\sin x\,+\,\cos x}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\cos (\pi/2 \,-\, x)\,+\,\cos x}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{2 \cos(\pi/4) \cos(\pi/4\,-\, x)}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\sqrt{2} \cos(\pi/4 \,-\, x)}}    \\
\end{array}}$                           

• Derivative of (π/4 −x) w.r.t x is −1.
• So we put u = (π/4 −x)
⇒ $\small{\frac{du}{dx}~=~-1}$
⇒ −dx = du
• Also, since u = (π/4 −x), we get: x = π/4 − u
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{(-1)(-1)\sin x}{\sqrt{2} \cos(\pi/4 \,-\, x)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-\sin (\pi/4 \,-\, u)}{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-\sin (\pi/4) \cos u ~+~\cos (\pi/4) \sin u}{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cos u ~+~(1/\sqrt 2)\sin u }{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) ~+~(1/\sqrt 2) \tan u }{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2)(1~-~\tan u)}{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1~-~\tan u)}{2}\right] \, du}}    \\
\end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{-1~+~\tan u}{2}\right] \, du}}    & {~=~}    &{\int{\left[\frac{-1}{2}\right] \, du}~+~\int{\left[\frac{\tan u}{2}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \int{\left[1 \right] \, du}~+~\frac{1}{2} \int{\left[\tan u \right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \left[u\,+\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos u |\,+\, C_2 \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\frac{1}{2} \left[(\pi/4)\,-\,x\,+\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,-\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|\cos (\pi/4) \, \cos x~+~\sin (\pi/4)  \,\sin x \right |\,+\, C_2 \right]}    \\
{~\color{magenta}    8    }    &{{}}   &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2) \, \cos x ~+~  (1/\sqrt 2) \,\sin x \right |\,+\, C_2 \right]}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2)(\cos x ~+~\sin x)  \right |\,+\, C_2 \right]}    \\
{~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log (1/\sqrt 2)~-~ \log \left|(\cos x ~-~\sin x)  \right |\,+\, C_2 \right]}    \\
{~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{C_3}{2}~-~\frac{\log \left|(\cos x ~+~\sin x)  \right |}{2}\,+\,\frac{C_4}{2}}    \\
{~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,-\,\frac{\log \left|(\cos x ~+~\sin x)  \right |}{2}\,+\,C}    \\
\end{array}}$                           
                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iv):
1. The given expression can be rearranged as follows:
$\small{\frac{1}{x\,+\, x \log x}~=~\frac{1}{x(1\,+\, \log x)}}$                     

• Derivative of (1+log x) w.r.t x is: (0+1/x) = 1/x.
• So we put u = 1+ log x
⇒ $\small{\frac{du}{dx}~=~\frac{1}{x}}$
⇒ $\small{du~=~\frac{1}{x} {dx}}$

2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{x\,+\, x \log x}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{1}{x(1\,+\, \log x)}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1}{u}\right] \, du}}    \\
\end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{u}\right] \, du}}    & {~=~}    &{\int{\left[\frac{1}{x(1\,+\, \log x)}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1}{u}\right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\log|(1\,+\, \log x)|~+~C}    \\
\end{array}}$

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The link below gives a few more solved examples:

Exercise 23.2

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

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23.4 - Standard Integrals of Trigonometric Functions

In the previous section, we saw integration by substitution. We saw some solved examples also. In this section, we will see some standard integrals of trigonometric functions.

In the table of integrals that we saw at the beginning of this chapter, there is the integral of sin x and also the integral of cos x. But altogether, there are six trigonometric functions. They are sine, cos, tan, csc, sec and cot. So there are four results missing from the table. This is because, integrals of sine and cosine can be obtained easily by inspection. The integrals of other four need to be derived using appropriate methods. We can use the method of substitution.

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First we will find $\small{\int{\left[\tan x \right]dx}}$
1. We have: $\small{\tan x ~=~\frac{\sin x}{\cos x}}$
• −sin x is the derivative of cos x
• So we put u = cos x
⇒ $\small{\frac{du}{dx}}$ = −sin x
⇒ −sin x dx = du

2. So we want:
$\small{\int{\left[\tan x \right]dx}}$
= $\small{\int{\left[\frac{\sin x}{\cos x} \right]dx}~=~\int{\left[\frac{(-1) \sin x}{(-1) \cos x} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1) \cos x} \right]du}~=~\int{\left[\frac{1}{(-1) u} \right]du}~=~(-1)\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{- \log |u| \,+\,C~=~-\log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~\log\left[\frac{1}{|u|}  \right]\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\tan x \right]dx~=~\log\left[\frac{1}{|\cos x|}  \right]\,+\,C~=~\log \left[|\sec x| \right]\,+\,C}}$

• We can write:
$\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$

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Next we will find $\small{\int{\left[\cot x \right]dx}}$
1. We have: $\small{\cot x ~=~\frac{\cos x}{\sin x}}$
• cos x is the derivative of sin x
• So we put u = sin x
⇒ $\small{\frac{du}{dx}}$ = cos x
⇒ cos x dx = du

2. So we want:
$\small{\int{\left[\cot x \right]dx}}$
= $\small{\int{\left[\frac{\cos x}{\sin x} \right]dx}}$
= $\small{\int{\left[\frac{1}{\sin x} \right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{\log[|u|] \,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$

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Next we will find $\small{\int{\left[\sec x \right]dx}}$
1. We have: $\small{\sec x ~=~\frac{\sec x(\sec x\,+\,\tan x)}{\sec x\,+\,\tan x}~=~\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x}}$
• (sec² x + sec x tan x) is the derivative of (tan x + sec x)
• So we put u = tan x + sec x
⇒ $\small{\frac{du}{dx}~=~\sec^2 x\,+\, \sec x \, \tan x}$
⇒ (sec² x + sec x tan x) dx = du

2. So we want:
$\small{\int{\left[\sec x \right]dx}}$
= $\small{\int{\left[\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x} \right]dx}}$
= $\small{\int{\left[\frac{1}{\sec x\,+\,\tan x} \right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{\log[|u|] \,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$

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Finally we will find $\small{\int{\left[\csc x \right]dx}}$
1. We have: $\small{\csc x ~=~\frac{\csc x(\csc x\,+\,\cot x)}{\csc x\,+\,\cot x}~=~\frac{\csc^2 x\,+\, \csc x \, \cot x}{\csc x\,+\,\cot x}}$
• (−csc² x − csc x cot x) is the derivative of (cot x + csc x)
• So we put u = cot x + csc x
⇒ $\small{\frac{du}{dx}~=~-\csc^2 x\,-\, \csc x \, \cot x}$
⇒ −(csc² x + csc x cot x) dx = du

2. So we want:
$\small{\int{\left[\csc x \right]dx}}$
= $\small{\int{\left[\frac{(-1)(\csc^2 x\,+\, \csc x \, \cot x)}{(-1)(\csc x\,+\,\cot x)} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1)(\csc x\,+\,\cot x)} \right]du}~=~(-1) \int{\left[\frac{1}{u} \right]du}}$

• This integration gives:
$\small{(-1) \log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~ \log[\frac{1}{|u|}] \,+\,C}$

3. Substituting for u, we get:
$\small{\log \left[\left|\frac{1}{\csc x\,+\,\cot x} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{\csc^2 x\,-\,\cot^2 x} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{1} \right| \right]\,+\,C~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$ 

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Let us compile the six results related to trigonometric ratios:
1. $\small{\int{\left[\sin x \right]dx~=~- \cos x \,+\,C}}$

2. $\small{\int{\left[\cos x \right]dx~=~ \sin x \,+\,C}}$

3. $\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$

4. $\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$

5. $\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$

6. $\small{\int{\left[\csc x \right]dx}~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$

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Now we will see some solved examples:

Solved example 23.6
Find the following integrals:
(i) $\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}}$

(ii) $\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}}$

(iii) $\small{\int{\left[\frac{1}{1\,+\, \tan x} \right]dx}}$

Solution:
Part (i):
1. −sin x is the derivative of cos x. There is no −ve sign in the question. But it can be adjusted by multiplying two −ve signs.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}~=~-\sin x}$
⇒ −sinx dx = du
• Also, when u = cos x, we get: sin²x = 1 − u²
2. So we want:
$\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}~=~\int{\left[(-1)(-1) \sin^3 x \, \cos^2 x \right]dx}}$

= $\small{\int{\left[(-1)\sin^2 x \,\, u^2 \right]du}~=~\int{\left[(-1)(1 - u^2) \, u^2 \right]du}}$
• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[(-1)(1 - u^2) \, u^2\right] \, du}}    & {~=~}    &{\int{\left[-u^2 + u^4 \, \right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[-u^2 \, \right] \, du}~+~\int{\left[u^4 \, \right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{- u^{3}}{3}\,+\,C_1 \right]~+~\left[\frac{u^{5}}{5} \,+\,C_2 \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{- u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C}    \\
\end{array}}$                           


3. Substituting for u, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x \, \cos^2 x\right] \, du}}    & {~=~}    &{\frac{-u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos^{3} x}{3}\,+\, \frac{\cos^{5} x}{5} \,+\, C}    \\
\end{array}}$

Part (ii):
1. The derivative of (x+a) is 1.
• So we put u = x+a
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
• Also, since u = x+a, we get: x = u−a
2. So we want:
$\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}~=~\int{\left[\frac{\sin x}{\sin(u)} \right]dx}}$

= $\small{\int{\left[\frac{\sin (u-a)}{\sin(u)} \right]du}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{\sin x}{\sin (x+a)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin (u-a)}{\sin u}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\sin u \cos a \,-\, \cos u \sin a }{\sin u}\right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\cos a \,-\, \cot u \sin a\right] \, du}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\cos a \right] \, du}~-~\int{\left[\cot u \sin a\right] \, du}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\cos a \int{\left[1 \right] \, du}~-~ \sin a \int{\left[\cot u \right] \, du}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\cos a \left[u \,+\,C_1 \right]~-~\sin a \left[\log |\sin u | \,+\,C_2 \right]}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{(\cos a)u \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin u | \,-\,(\sin a) C_2}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{(\cos a)(x+a) \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,+\, a \cos a \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2}    \\
{~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,+\, C_3 \,+\,C_4 ~-~(\sin a)\log |\sin (x+a) | \,-\,C_5}    \\
{~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,-\, (\sin a)\log |\sin (x+a) | \,+\,C}    \\
\end{array}}$

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.                            

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,+\,\tan x}}    & {~=~}    &{\frac{1}{1\,+\,(\sin x / \cos x)}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin x\,+\,\cos x}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin x\,+\,\sin(\pi/2 \,-\, x)}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{2 \sin(\pi/4) \cos(x\,-\, \pi/4)}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}}    \\
\end{array}}$                           

• Derivative of (x−π/4) w.r.t x is 1.
• So we put u = (x−π/4)
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
• Also, since u = (x−π/4), we get: x = u + π/4
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\cos (u\,+\, \pi/4)}{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos u \cos (π/4)~-~\sin u \sin(π/4)}{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos u (1/\sqrt 2)~-~\sin u (1/\sqrt 2)}{\sqrt{2} \cos u}\right] \, du}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{(1/\sqrt 2)~-~\tan u (1/\sqrt 2)}{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{(1/\sqrt 2)(1~-~\tan u)}{\sqrt{2}}\right] \, du}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1~-~\tan u}{2}\right] \, du}}    \\
\end{array}}$                           

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1~-~\tan u}{2}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{1}{2}\right] \, du}~-~\int{\left[\frac{\tan u}{2}\right] \, du}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[1 \right] \, du}~-~\frac{1}{2} \int{\left[\tan u \right] \, du}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[u\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec u |\,+\, C_2 \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec (x\,-\,(\pi/4)) |\,+\, C_2 \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log |\sec (x\,-\,(\pi/4)) |\,-\, C_2 \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log \left|\frac{1}{\cos (x\,-\,(\pi/4))} \right |\,-\, C_2 \right]}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\left(\frac{1}{\cos (x\,-\,(\pi/4))}\right)^{-1} \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos (x\,-\,(\pi/4)) \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, \cos(\pi/4) ~+~\sin x  \,\sin(\pi/4) \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, (1/\sqrt 2) ~+~\sin x  \,(1/\sqrt 2) \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(1/\sqrt 2)(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log (1/\sqrt 2)~+~ \log \left|(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\
{~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(\cos x ~+~\sin x)  \right |\,+\, C_5 \right]}    \\
{~\color{magenta}    {14}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2} \,+\, \frac{1}{2} \left[\log \left|(\cos x ~+~\sin x)  \right | \right]\,+\, C}    \\
\end{array}}$                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

---

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

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23.3 - Methods of Integration

In the previous section, we completed a discussion on integration by method of inspection. This method is not applicable in all cases. So we need to learn about some other methods also. Three prominent methods are:
    ♦ Integration by substitution
    ♦ Integration using partial fractions
    ♦ Integration by parts
• In this section, we will see integration by substitution.

• Recall the chain rule that we used for differentiating functions of the form f(g(x)).
For example:
$\small{\frac{d}{dx}\left[\sin (2x+3)  \right]~=~\cos(2x+3) \times \frac{d}{dx} \left[2x+3 \right]~=~2 \cos(2x+3)}$
• So if the integrand is 2 cos(2x+3), we can use a reverse process to get an integral sin(2x+3)

The reverse process can be explained in 6 steps:
1. We want: $\int{\left[2 \cos (2x+3) \right] dx}$  
2. The integrand is 2 cos(2x+3). We closely examine the integrand and find two parts such that, one of them is the derivative of the other.
• In this case, the two parts are: “2” and “(2x + 3)”
    ♦ “2” is the derivative of “(2x + 3)”
3. We put u = 2x+3
Then $\small{\frac{du}{dx}}$ = 2, which gives 2 dx = du
4. So the expression in (1) can be rewritten as:
$\small{\int{\left[\cos (u) \right] du}}$
• Note that,
    ♦ 2x+3 is replaced by u
    ♦ 2dx is replaced by du
5. Integration of the expression in (4) gives: sin u + C
• Replacing u, we get: sin(2x+3) + C
6. This method is called integration by substitution.
Note that, we did the substitution of (2x+3) by u.

Another example can be written in 6 steps:
1. We want: $\small{\int{\left[6x(3x^2 \,+\, 4)^4 \right] dx}}$  
2. The integrand is 6x (3x² + 4)⁴. We closely examine the integrand and find two parts such that, one of them is the derivative of the other.
• In this case, the two parts are: “6x” and “(3x² + 4)”
    ♦ “6x” is the derivative of “(3x² + 4)”
3. We put u = 3x² + 4
Then $\small{\frac{du}{dx}}$ = 6x, which gives 6x dx = du
4. So the expression in (1) can be rewritten as:
$\int{\left[u^4 \right] du}$  
• Note that,
    ♦ 3x² + 4 is replaced by u
    ♦ 6x dx is replaced by du
5. Integration of the expression in (4) gives: $\small{\frac{u^5}{5} \,+\, C}$
Replacing u, we get: $\small{\frac{(3x^2 \,+\, 4)^5}{5} \,+\, C}$
6. So we used integration by substitution.
Note that, we did the substitution of (3x² + 4) by u.

---

Now we will see some solved examples

Solved example 23.5
Use substitution to find:
(i) $\small{\int{\left[3 x^2 (x^3 \,-\,3)^2 \right]dx}}$

(ii) $\small{\int{\left[\sin(mx) \right]dx}}$

(iii) $\small{\int{\left[\frac{\tan^4 \sqrt{x} \, \sec^2 \sqrt{x}}{\sqrt{x}} \right]dx}}$

(iv) $\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}}$

(v) $\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}}$

(vi) $\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right]dx}}$

Solution:
Part (i):
1. 3x² is the derivative of (x³ - 3)
• So we put u = (x³ - 3)
⇒ $\small{\frac{du}{dx}}$ = 3x²
⇒ 3x² dx = du

2. So we want:
$\small{\int{\left[(u)^2 \right]du}}$
• This integration gives $\small{\frac{u^3}{3}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[3 x^2 (x^3 \,-\,3)^2 \right]dx~=~\int{\left[u^2 \right]du}~=~\frac{u^3}{3}\,+\,C~=~\frac{(x^3 \,-\, 3)^3}{3}\,+\,C}}$

Part (ii):
1. m is the derivative of mx. There is no independent m in the question. But since m is a constant, there will not be any issue.
• So we put u = mx
⇒ $\small{\frac{du}{dx}}$ = m
⇒ m dx = du

2. So we want:
$\small{\int{\left[\frac{m \sin(mx)}{m} \right]dx}}$
= $\small{\int{\left[\frac{\sin(u)}{m} \right]du}}$
• This integration gives $\small{\frac{- \cos(u)}{m}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\sin(mx) \right]dx}~=~\frac{- \cos(mx)}{m }\,+\,C}$

Part (iii):
1. $\small{\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}}}$ is the derivative of $\small{\tan \sqrt{x}}$. There is no independent '2' in the question. But since '2' is a constant, there will not be any issue.
• So we put u = $\small{\tan \sqrt{x}}$
⇒ $\small{\frac{du}{dx}}~=~\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}}$

⇒ $\small{\left(\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}} \right)dx~=~du}$

2. So we want:
$\small{\int{\left[2 \tan^4 \sqrt{x} \,\left(\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}} \right) \right]dx}~=~\int{\left[2\, u^4 \right]du}}$
• This integration gives $\small{\frac{2\,u^5}{5}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\tan^4 \sqrt{x} \, \sec^2 \sqrt{x}}{\sqrt{x}} \right]dx}~=~\frac{2\, \tan^5 \sqrt{x}}{5}\,+\,C}$

Part (iv):
1. $\small{\frac{1}{1\,+\,x^2}}$ is the derivative of $\small{\tan^{-1} x}$.
• So we put u = $\small{\tan^{-1} x}$
⇒ $\small{\frac{du}{dx}~=~\frac{1}{1\,+\,x^2}}$

⇒ $\small{\left(\frac{1}{1\,+\,x^2} \right)dx~=~du}$

2. So we want:
$\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}~=~\int{\left[\sin(u) \right]du}}$
• This integration gives $\small{- \cos(u)\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}~=~- \cos(\tan^{-1} x)\,+\,C}$

Part (v):
1. −sin x is the derivative of cos x. There is no −ve sign in the question. But since (−1) is a constant, there will not be any issue.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}}$ = −sin x
⇒ −sin x dx = du

2. So we want:
$\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}}$
= $\small{\int{\left[\frac{(-1) \sin x}{(-1) \cos^3 x} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1) u^3} \right]du}}$

• This integration gives $\small{(-1) \frac{1}{(-2) u^2}\,+\,C~=~ \frac{1}{2 u^2}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}~=~ \frac{1}{2 \cos^2 x}\,+\,C}$

Part (vi):
1. The derivative of (x − 1) is 1. The number '1' will be already available.
• So we put u = x − 1
⇒ $\small{\frac{du}{dx}}$ = 1
⇒ (1) dx = du
• Also, u = x − 1 gives: x = u + 1

2. So we want:
$\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right](1)dx}}$
= $\small{\int{\left[\frac{u\,+\,1}{\sqrt u} \right]du}}$

• This integration can be done as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{u\,+\,1}{\sqrt u}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{u\,+\,1}{\sqrt u} \right] \, dx}~=~\int{\left[\frac{u}{\sqrt u} ~+~\frac{1}{\sqrt u} \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{u^{1/2}}{1} ~+~\frac{u^{-1/2}}{1} \right] \, dx}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{u^{1/2}}{1}  \right] \, dx}~+~\int{\left[\frac{u^{-1/2}}{1} \right] \, dx}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{u^{3/2}}{3/2}\,+\,C_1 \right]~+~\left[\frac{u^{1/2}}{1/2} \,+\,C_2 \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{u^{3/2}}{3/2}\,+\, \frac{u^{1/2}}{1/2} \,+\, C}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 u^{3/2}}{3}\,+\, \frac{2 u^{1/2}}{1} \,+\, C}    \\
\end{array}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right]dx}~=~ \frac{2 (x\,-\,1)^{3/2}}{3}\,+\, \frac{2 (x\,-\,1)^{1/2}}{1} \,+\, C}$

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In the next section, we will see some standard integrals of trigonometric functions.

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23.2 - Solved Examples on Method by Inspection

In the previous section, we saw integration by method of inspection. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.2
Find the following integrals:
$(i)~\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~~~~(ii)~\int  {\left[x^{2/3} \,+\,1\right]dx}$

$(iii)~\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x}\right] \, dx}~~~~(iv)~\int{\left[\frac{4}{1+x^2}\right] \, dx}$

Solution:
Part (i):
1. We want $\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}$
• This can be written as:
$\int{\left[x~-~\frac{1}{x^2} \right] \, dx}$
⇒ $\int{\left[x~-~x^{-2} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}$
• We search for F and find that:
    ♦ x, is the derivative of $\frac{x^2}{2}$
    ♦ x⁻², is the derivative of $\frac{x^{-1}}{-1}$ which is −(1/x)
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}}    & {~=~}    &{\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^2}{2}\,+\,C_1 \right] ~-~\left[\frac{-1}{x}\,+\,C_2 \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C_1 \,-\,C_2}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C}    \\
\end{array}$

• This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~=~\frac{x^2}{2}\,+\, \frac{1}{x}$ is an anti derivative of $\frac{x^3 - 1}{x^2}$. 

Part (ii):
1. We want $\int{\left[x^{2/3} \,+\,1 \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}$
• We search for F and find that:
    ♦ x^2/3, is the derivative of $\frac{x^{5/3}}{5/3}$
    ♦ 1, is the derivative of x
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[x^{2/3} \,+\,1 \right] \, dx}}    & {~=~}    &{\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^{5/3}}{5/3}\,+\,C_1 \right] ~+~\left[x\,+\,C_2 \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C_1 \,+\,C_2}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C}    \\
\end{array}$                           

• This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[x^{2/3} \,+\,1 \right] \, dx}~=~\frac{3 x^{5/3}}{5}\,+\, x$ is an anti derivative of $x^{2/3} \,+\,1$.

Part (iii):
1. We want $\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{3/2} \right]\,dx}~+~\int{\left[2 e^x \right]}~-~\int{\left[\frac{1}{x} \right] \, dx}$
• We search for F and find that:
    ♦ x^3/2, is the derivative of $\frac{x^{5/2}}{5/2}$
    ♦ 2e^x, is the derivative of 2e^x
    ♦ 1/x, is the derivative of log|x|
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}}    & {~=~}    &{\int{\left[x^{3/2} \right] \, dx}~+~\int{\left[2 e^x  \right] \, dx}~-~\int{\left[\frac{1}{x} \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^{5/2}}{5/2}\,+\,C_1 \right]~+~\left[2 e^x \,+\,C_2 \right]~-~\left[\log|x| \,+\, C_3\right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C_1 \,+\,C_2 \,-\,C_3}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C}    \\
\end{array}$                           

• This is the general form of the required anti derivative.
◼ Remarks:
3 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}~=~\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x|$ is an anti derivative of $x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x}$.

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Once we understand the basics, there is no need to write all the detailed steps. So from now on wards, we shall write only the necessary steps.

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Solved example 23.3
Find the following integrals:
$(i)~\int{\left[\sin x \,+\, \cos x \right] \, dx}~~~~~~~(ii)~\int  {\left[\csc x (\csc x \,+\, \cot x)\right]dx}~~~~~~~(iii)~\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin x \,+\, \cos x \right] \, dx}}    & {~=~}    &{\int{\left[\sin x  \right] \, dx}~+~\int{\left[\cos x \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[- \cos x\,+\,C_1 \right]~+~\left[\sin x \,+\,C_2 \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{-\cos x \,+\, \sin x \,+\, C}    \\
\end{array}$

Part (ii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\csc x (\csc x \,+\, \cot x) \right] \, dx}}    & {~=~}    &{\int{\left[\csc^2 x \,+\, \csc x \cot x \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\csc^2 x \right] \, dx}~+~\int{\left[\csc x \cot x \right] \, dx}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[-\cot x \,+\, C_1 \right]~+~\left[-\csc x \,+\,C_2 \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\cot x \,-\, \csc x \,+\,C}    \\
\end{array}$

Part (iii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}}    & {~=~}    &{\int{\left[\frac{1 }{\cos^2 x} ~-~\frac{\sin x}{\cos^2 x}\right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sec^2 x ~-~ \sec x \tan x \right] \, dx}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sec^2 x \right] \, dx}~-~\int{\left[\sec x \tan x \right] \, dx}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left[\tan x \,+\, C_1 \right]~-~\left[\sec x \,+\,C_2 \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\tan x \,-\, \sec x \,+\,C}    \\
\end{array}$

Part (iv):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{4}{1+x^2} \right] \, dx}}    & {~=~}    &{4 \int{\left[\frac{1}{1+x^2} \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{4 \left[\tan^{-1} x \,+\, C \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4 \tan^{-1} x \,+\, C}    \\
\end{array}$                           
 

Solved example 23.4
Find the anti derivative F of f defined by f (x) = 4x³ − 6, where F(0) = 3
Solution:
1. First we will write the general form of the anti derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[4x^3 \,-\,6 \right] \, dx}}    & {~=~}    &{\int{\left[4 x^3 \right] \, dx}~-~\int{\left[6 \right] \, dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{4 x^4}{4} \,+\, C_1 \right]~-~\left[6x \,+\,C_2 \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{x^4 \,-\, 6x \,+\,C}    \\
\end{array}$

• So the general form of the anti derivative is: x⁴ − 6x + C

2.Given that, when x = 0, the anti derivative becomes 3.
• So we can write: x⁴ − 6x + C = 3
⇒ (0)⁴ − 6(0) + C = 3
⇒ C = 3

3. So based on the general form, we can write:
The exact anti derivative is: x⁴ − 6x + 3

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The link below gives a few more solved examples:

Exercise 23.1

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In the next section, we will see Methods of Integration.

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