Integration by Substitution Method

Integration by substitution is a method used to simplify an integral by replacing a complex expression with a new variable (u), making the integration process easier.

The diagram below illustrates the substitution process used to simplify an integral.

For example:

∫ 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)

Using the Substitution Method

Integration by substitution (also called u-substitution) is used when an integral contains a function and its derivative. It helps simplify a complex integral by replacing part of the expression with a new variable.

This method is commonly used when the integral is of the form:

∫ f(g(x)) · g'(x) dx

We let: u = g(x)

Then: du = g'(x) dx

Substituting these into the integral gives:

∫ f(u) du.

This new integral is usually easier to solve. After integrating, we replace u with g(x) to express the answer in terms of x.

In simple terms, integration by substitution is used whenever a change of variable can make the integral easier to evaluate

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

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

Solution:

Let, I =  ∫ 2x sin(x² − 5) dx..(i)

Here, f = sin, g(x) = x² - 5, g'(x) = 2x

Substituting, x² - 5 = t

Differentiating the above equation

2x 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 (x² - 5) + c

This is the required solution for given integration.

Example 4: 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 5: 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 6: 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.

Practice Questions

1. \int 2x \cos(x^2) \, dx

2. \int \sin(3x) \cos(3x) \, dx

3. \int x e^{x^2} \, dx

4. \int \frac{2x}{x^2 + 1} \, dx

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

6. \int (3x^2 - 5)^5 \cdot 6x \, dx

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• Indefinite Integral
• Integration Formulas
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