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.

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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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