Integration by Substitution Method

Last Updated : 02 Jan, 2025

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

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.

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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

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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. [Tex]\int 2x \cos(x^2) \, dx[/Tex]

2. [Tex]\int \sin(3x) \cos(3x) \, dx[/Tex]

3. [Tex]\int x e^{x^2} \, dx[/Tex]

4. [Tex]\int \frac{2x}{x^2 + 1} \, dx[/Tex]

5. [Tex]\int \frac{\sin(x)}{\cos^3(x)} \, dx[/Tex]

6.[Tex] \int (3x^2 – 5)^5 \cdot 6x \, dx[/Tex]

Integration by Partial Fractions

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Article Tags :

Integration by Substitution Method

Substitution Method of Integration

When to use Integration by Substitution?

Steps to Integration by Substitution

Integration by Special Substitution

Integration by Substitution Examples

Integration by Substitution Questions

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Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications of Derivatives

Chapter 7: Integrals

Chapter 8: Applications of Integrals

Chapter 9: Differential Equations