Integration by Substitution Method (original) (raw)
Last Updated : 2 Jun, 2026
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
- \int (3x^2 - 5)^5 \cdot 6x \, dx