# Integration by Substitution

• Difficulty Level : Medium
• Last Updated : 05 Mar, 2021

Integration is a method of finding a function f(x) whose derivative Df(x) is equal to the given function. That is why it’s also known as anti-derivative. The integral is usually calculated to find the function which provides information about the area, displacement, volume, which appears due to the collection of small data which can’t be measured singularly. Here we can define the integration by substitution method or Integration of the form f(p)p'(x).

1. Integration by substitution method
2. Integration of the form f(p)p'(x)

### Integration by Substitution Method

Integration by substitution method can be used whenever the given function f(x), and is multiplied by the derivative of given function f(x)’, i.e. of this form ∫g( f(x) f(x)’ ) dx. When the function that is to be integrated is not in a standard form it can sometimes be transformed to integrable form by a suitable substitution. The integral

∫ f {g (x)} g’ (x) dx can be converted to

∫f (θ) dθ by substituting g (x) by θ, so that if

∫f (θ) dθ = F(θ) + c, Then

∫f{g (x)} g’ (x) dx = F{g(x)} + c

This is a direct consequence of chain rule, for

d/dx [F {g(x)} + c] = d/dθ [F(θ) + c] . dθ/dx = f {g(x)} g’(x)

There is no definite formula for substitution. Keen observation of the form of the integrand will help in choosing the function for which substitution is to be made. However, one must be sure that the derivative of the function so chosen must be present along with dx as in the above case. Occasionally mere adjustment of a constant may be necessary. Any symbol for variable viz. s, t, u, v, w, x, y, z may be chosen for substitution other than the variable of the given integral. However, after the integration is over, the original variable should be put back.

### Examples

Example 1: Integrate ∫ 2x.cos (x2) dx

Solution:

Let, I = ∫ 2x. cos (x2) dx                     ………….                                            (i)

Substituting  x2 = t                            …………                                            (ii)

by differentiate the above equation

2x dx = dt                             …………                                             (iii)

put the equ (ii) and (iii) in equ (i), we get

= ∫ cos t dt

Integrate the above equation then, we get

=  sin t + c

Put the value of t in the above equ

= sin (x2) + c

Hence, I = sin (x2) + c

Example 2: Integrate ∫ sin (x3) . 3x2 dx

Solution:

Let, I = ∫ sin (x3). 3x2 dx                            ………….                                            (i)

Substituting  x3 = t                                    …………                                            (ii)

by differentiate the above equation

3x2 dx = dt                                    …………                                             (iii)

put the equ (ii) and (iii) in equ (i), we get

= ∫ sin t dt

Integrate the above equation then, we get

=  – cos t + c

Put the value of t in the above equ

= – cos (x3) + c

Hence, I = – cos (x3) + c

Example 3: Integrate ∫ 2x cos(x2 − 5) dx

Solution:

Let, I =  ∫ 2x cos(x2 − 5) dx                      ………                          (i)

Put, x2 – 5 = t                                          ………                          (ii)

differentiate the above equ

2x dx = dt                                   ………                           (iii)

put the equ (ii) and (iii) in equ (i), we get

= ∫ cos (t) dt

Integrate the above equ then, we get

= sin t + c

Put the value of t in above equ

= sin (x2 – 5) + c

Hence, I = sin (x2 – 5) + c

Example 4: Integrate ∫x/(x2 + 1) dx

Solution:

Let, I = ∫ x / (x2+1) dx

we rearrange the above equ

= (1/2) ∫ 2x / (x2+1) dx                                   (i)

put ,   x2 + 1 = t                                                  (ii)

2x dx = dt                                                    (iii)

put the equ (ii) and (iii) in equ (i), we get

= (1/2) ∫ 1/t  dt

Integrate the above equ then, we get

= (1/2) log t + c

now, put the value of t in the above equ

= (1/2) log (x2 +1) + c

Hence, I = (1/2) log (x2 + 1) + c

Example 5: Integrate ∫ (2x + 3) (x2 + 3x)2 dx

Solution:

Let, I = ∫ (2x + 3) (x2 + 3x)2 dx                                               (i)

Substitute x2 + 3x = t                                                          (ii)

differenitate the above equ

2x + 3 dx = dt                                                    (iii)

put the equ (ii) and (iii) in equ (i), we get

= ∫ t2 dt

Integrate the above equ then, we get

= t3/3 + c

now, put the value of t in the above equ

= (x2 + 3x)3 / 3 + c

Hence, I = (x2 + 3x)3/3 + c

### Integration of the form f(p)p'(x)

f (p). p'(x) where p is a function of x

I = ∫ f(p). p'(x) dx

Let p(x) = t

p'(x) dx = dt

I = ∫f(t) dt

### Examples

Example 1: ∫cos(x2) 2x dx

Solution:

∫cos(x2) 2x dx

Here, f = cos, g(x) = x2, g'(x) = 2x

So put,

x2 = t  …..         (1)

differentiate the above equ w.r.t x

2x dx = dt   …..     (2)

so the equ is,

∫cost dt   …..        (3)

from integrate the equ (3)

sin t + c   ……        (4)

then put the value t in equ (4)

The final answer is = sin(x2) + c

Example 2: Integrate ∫ cos (x3) . 3x2 dx

Solution:

Let, I = ∫ cos (x3). 3x2 dx                                                                        (i)

Here, f = cos,  g(x) = x3,  g'(x) = 3x2

Substituting  x3 = t                                                                                (ii)

by differentiate the above equation

3x2 dx = dt                                                                          (iii)

put the equ (ii) and (iii) in equ (i), we get

= ∫ cos t dt

Integrate the above equation then, we get

=  sin t + c

Put the value of t in the above equ

= sin (x3) + c

Hence, I = sin (x3) + c

Example 3: Integrate ∫ 2x sin(x2 − 5) dx

Solution:

Let, I =  ∫ 2x cos(x2 − 5) dx                                              (i)

Here, f= cos,  g(x) = x2 – 5,  g'(x) = 2x

Put, x2 – 5 = t                                                                 (ii)

differentiate the above equ

2x dx = dt                                                               (iii)

Put the equ (ii) and (iii) in equ (i), we get

= ∫ cos (t) dt

Integrate the above equ then, we get

= – sin t + c

Put the value of t in above equ

= – sin (x2 – 5) + c

Hence, I = – sin (x2 – 5) + c

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