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 is one of the important methods for finding the integration of the function where direct integration can not be easily found. This method is very useful in finding the integration of complex functions. We use integration by substitution to reduce the given function into the simplest form such that its integration is easily found.

In calculus, integration by substitution is a method used to solve integrals and antiderivatives, also referred to as u-substitution, the reverse chain rule, or change of variables. It serves as the analog to the chain rule used in differentiation. Essentially, this method can be viewed as the chain rule in reverse to simplify and evaluate integrals.

Now let’s learn more about the integration by substitution method, Integration by Substitution examples, and others in this article.