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Integration by Substitution Method – Formula, Examples & Questions, Class 12 Math Notes Study Material Download Free PDF

Integration by substitution or u-substitution is a highly used method of finding the integration of a complex function by reducing it to a simpler function and then finding its integration.

Suppose we have to find the integration of f(x) where the direct integration of f(x) is not possible. So we substitute x = g(t).

For, I = ∫f(x).dx

substituting x = g(t) 

⇒ dx/dt = g'(t)

⇒ dx = g'(t).dt

Thus, I = ∫f(x).dx = ∫f(g(t)).g'(t).dt

This integral becomes easy to calculate.

Integration by Substitution Method

Integration by Substitution

Substitution Method of Integration

Integration by substitution method can be used whenever the given function f(x) and its derivative f'(x) are multiplied and given as a single function i.e. the given function is of form ∫g(f(x) f(x)’ ) dx then we use integration by substitution method.

Sometimes the given function is not in the form where we can directly apply the  Substitution Method then we transform the function into such a form where we can use the Substitution Method.

For example, we can use

∫ f {g (x)} g’ (x) dx can be converted to another form ∫f(θ) dθ, By substituting g (x) with θ, 

Such that,

∫f (θ) dθ = F(θ) + c, 

Then

∫f{g (x)} g’ (x) dx = F{g(x)} + c

This can be proved using the chain rule, as follows

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

There is no direct method of substitution we have to observe the function carefully and then have to decide what is to be substituted in the function to make it easily integrable.

When to use Integration by Substitution?

Integration by substitution is a widely used method for solving the integration of complex functions. It is also called the “Reverse Chain Rule”. We use this method when the given integral is in the form,

∫ f(g(x)).g'(x).dx

We can transform the given integral n this form. We substitute 

g(x) = u 

differentiating both sides we get,

g'(x).dx = du

Substituting this in the given function.

∫ f(g(x)).g'(x).dx = ∫ f(u).du

Thus the ∫ f(g(x)).g'(x).dx is easily integrable in form ∫ f(u).du. After this, we substitute back the value of u = g(x) to get the final solution.

Steps to Integration by Substitution

Integration by Substitution is achieved by following the steps discussed below,

  • Step 1: Choose the part of the function (say g(x)) as t which is to be substituted.
  • Step 2: Differentiate the equation g(x) = t to get the value of d(t), here the value is dt = g'(x) dx
  • Step 3: Substitute the value of t and d(t) in the given function. Now the function becomes integrable.
  • Step 4: Integrate the reduce function to get the solution. 
  • Step 5: Substitute the value of t = g(x) in the final solution, to get the final answer.

Integration by Special Substitution

Various integration can be achieved by using the integration by substitution method. Some of the common forms of integrations that can be easily solved using the Integration by Substitution method are,

  • If the given function is in form f(√(a2 – x2)) we use substitution as, x = a sin θ or x = a cos θ
  • If the given function is in form f(√(x2 – a2)) we use substitution as, x = a sec θ or x = a cosec θ
  • If the given function is in form f(√(x2 + a2)) we use substitution as, x = a tan θ or x = a cot θ

Integration by Substitution Examples

Example 1: Integrate ∫ 2x.cos (x2) dx

Solution:

Let, I = ∫ 2x. cos (x2) dx . . . (i)

Substituting  x2 = t 

Differentiating the above equation

2x dx = dt

Substituting this in eq (i)

I = ∫ cos t dt

Integrating the above equation

I =  sin t + c

Putting back the value of t

I = sin (x2) + c

This is the required solution for given integration.

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

Solution:

Let, I = ∫ sin (x3). 3x2 dx . . . (i)

Substituting  x3 = t

Differentiating the above equation

3x2 dx = dt

Substituting this in eq (i)

I = ∫ sin t dt

Integrating the above equation

I =  – cos t + c 

Putting back the value of t

I = – cos (x3) + c

This is the required solution for given integration.

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

Solution:

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

Substituting x2 – 5 = t

Differentiating the above equation

2x dx = dt

Substituting this in eq (i)

I = ∫ cos (t) dt

Integrating the above equation

I = sin t + c

Putting back the value of t

I = sin (x2 – 5) + c

This is the required solution for given integration.

Example 4: Integrate ∫x/(x2 + 1) dx

Solution:

Let, I = ∫ x / (x2+1) dx

Rearranging the above equation

I = (1/2) ∫ 2x / (x2+1) dx . . . (i)

Substituting x2 + 1 = t

Differentiating the above equation

2x dx = dt

Substituting this in eq (i)

I = (1/2) ∫ 1/t  dt

Integrating the above equation

I = (1/2) log t + c

Putting back the value of t

I = (1/2) log (x2 +1) + c

This is the required solution for given integration.

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

Solution:

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

Substitute x2 + 3x = t

Differentiating the above equation

2x + 3 dx = dt

Substituting this in eq (i)

I = ∫ t2 dt  

Integrating the above equation

I = t3/3 + c

Putting back the value of t

I = (x2 + 3x)3 / 3 + c

This is the required solution for given integration.

Example 6: ∫cos(x2) 2x dx

Solution:

Let, I = ∫cos(x2) 2x dx . . . (i)

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

Substitute, x2 = t

Differentiating the above equation

2x dx = dt

Substituting this in eq (i)

I = ∫cost dt

Integrating the above equation

I = sin t + c

Putting back the value of t

I = sin(x2) + c

This is the required solution for given integration.

Example 7: 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

Differentiating the above equation

3x2 dx = dt

Substituting this in eq (i)

I = ∫ cos t dt

Integrating the above equation

I =  sin t + c

Putting back the value of t

I = sin (x3) + c

This is the required solution for given integration.

Example 8: 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

Substituting, x2 – 5 = t

Differentiating the above equation

2x dx = dt

Substituting this in eq (i)

I = ∫ cos (t) dt

Integrating the above equation

I = – sin t + c

Putting back the value of t

I = – sin (x2 – 5) + c

This is the required solution for given integration.

Integration by Substitution Questions

1. ∫2xcos⁡(x2)dx∫2xcos(x2)dx

2. ∫sin⁡(3x)cos⁡(3x)dx∫sin(3x)cos(3x)dx

3. ∫xex2dx∫xex2dx

4. ∫2xx2+1dx∫x2+12x​dx

5. ∫sin⁡(x)cos⁡3(x)dx∫cos3(x)sin(x)​dx

6.∫(3x2–5)5⋅6xdx∫(3x2–5)5⋅6xdx

Integration by Substitution – FAQs

What is Integration by Substitution?

Integration by substitution is the method of finding the integration of the complex function where the normal techniques of integration fail. In this method, we find the substitute part of the function with another function and then get the transformed function which can be easily integrated.

How to Integrate by Substitution?

We can easily integrate any function of the form,

∫ f(g(x)).g'(x).dx

by just simply substituting g(x) = t and then replacing g'(x).dx = dt. So the function becomes ∫f(t)dt which can be easily integrated.

Why we use the substitution method?

We use integration by substitution to find the solution of the given function where the normal integration techniques fail.

How to know when to use Integration by Substitution?

We use integration by substitution when the given integration is of form ∫ f(g(x)).g'(x).dx or can be easily changed to the above form.

What are some situations where Integration by Substitution is used?

Some conditions where we use integration by substitution are,

  • If the given function is in form f(√(a2 – x2)) we use substitution as, x = a sin θ or x = a cos θ
  • If the given function is in form f(√(x2 + a2)) we use substitution as, x = a tan θ or x = a cot θ
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