Integration By Substitution

Integration is a crucial topic in calculus, and one of the most powerful techniques for solving integrals is integration by substitution. This method allows us to simplify complex integrals into more manageable forms by making a substitution that makes the integral easier to solve.

1.0What is Integration by Substitution?

Integration by substitution is a technique that involves substituting a part of the integral with a new variable to simplify the expression. It’s similar to the chain rule in differentiation but in reverse. The goal is to make the integral easier to solve by changing the variables.

The idea behind substitution is to find a part of the integrand (the function being integrated) that can be replaced with a new variable, such that the new integral is simpler.

2.0Why Use Integration by Substitution?

• Simplification: It makes complex integrals easier to handle.
• Applicability: It works well when the integral has a composite function (like a function inside another function).
• Efficiency: It helps in solving integrals that might otherwise seem impossible.

3.0Integration by Substitution Formula

The general integration by substitution formula is as follows:

$\int f(g(x)).g^{\prime }(x)dx=\int f(u)du$

Where:

• u = g(x) (the substitution),
• du = g′(x) dx (the differential of u).

4.0Steps for Integration by Substitution

1. Identify the substitution: Choose a part of the integrand that can be replaced by a single variable, usually a function inside another function.
2. Compute du: Differentiate the chosen substitution to find du, and express dx in terms of du.
3. Substitute and simplify: Replace all occurrences of x with the new variable u and simplify the integral.
4. Integrate: Perform the integration with respect to u.
5. Substitute back: Once the integral is solved in terms of u, substitute back the original expression for u to get the answer in terms of x.

5.0Example of Integration by Substitution

Let’s solve an example problem using the integration by substitution method.

Question:$\int 2x.e^{x^2}dx$

Solution:

1. Choose the substitution: Here, we see that $x^2$ appears in the exponent of the exponential function. Let’s set: $u=x^2$
2. Compute du: Differentiate $u=x^2$ with respect to x:

$du=2xdx$

Therefore, 2x dx = du.

3. Substitute and simplify: Now, replace $x^2$ with u and 2x dx with du in the integral:

$\int 2x.e^{x^2}dx=\int e^{u}du$

4. Integrate: The integral of $e^u$ with respect to u is simply:

$\int e^{u}du=e^u+C$

Where C is the constant of integration.

5. Substitute back: Finally, replace u with $x^2$ to get the result in terms of xx:

  $e^u+C=e^{x^2}+C$

Thus, the solution to the integral is: $\int 2x.e^{x^2}dc=e^{x^2}+C$

6.0Solved Examples of Integration by Substitution

Example 1: $\int \frac{ln{x}^2}{x}dx$

Solution:

Let $ln{x}^2=t$

$\frac{1}{x^2}2xdx=dt$

$\frac{2dx}{x}=dt$

$\Rightarrow \int \frac{t}{2}dt=\frac{t^2}{4}+C$

$=\frac{(ln{x}^2{)}^2}{4}+C$

Example 2: $\int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx$

Solution :

Let $si{n}^{-1}x=t$

$\int \frac{1}{\sqrt{1-x^2}}dx=dt$

$\Rightarrow \int tdt=\frac{t^2}{2}+C$

$\Rightarrow (Si{n}^{-1}x{)}^2+C$

Example 3: $\int \frac{sin(ta{n}^{-1})}{1+x^2}dx$

Solution :

Put, x =t

$\frac{1}{1+x^2}dx=dt$

$\Rightarrow \int sintdt=-cost+C$

$=-cos(tan-1x)+C$

Example 4: $\int (4x+6)\sqrt{2x^2+6x+5}dx$

Solution:

Let $2x^2+6x+5=t$

$(4x+6)dx=dt$

$\Rightarrow \int \sqrt{t}dt=\frac{t^{3/2}}{\frac{3}{2}}+C$

$=\frac{2}{3}(2x^2+6x+5{)}^{\frac{3}{2}}+C$

Example 5: $\int \frac{\sqrt{tanx}}{sin2x}dx$

Solution:

$\int \frac{\sqrt{tanx}}{2tanx}(1+ta{n}^{2}x)dx$

$\Rightarrow \int \frac{\sqrt{tanx}}{2tanx}se{c}^{2}xdx$

Put $tanx=t$

$se{c}^{2}xdx=dt$

$\frac{1}{2}\int (t{)}^{-\frac{1}{2}}dt=\frac{t^{\frac{1}{2}}\times 2}{2\times 1}+C$

$=\sqrt{t}+C$

$=\sqrt{tanx}+C$    

Example 6: $\int \frac{2x^{3}dx}{1+x^2}$

Solution:

Put $1+x^2=t$

$2xdx=dt$

$\Rightarrow \int \frac{2x.x^{2}dx}{1+x^2}=\int \frac{(t-1)}{t}dt$

$=\int (1-\frac{1}{t})dt$

$=t-lnt+C$

$=(1+x^2)-ln(1+x^2)+C$

Example 7: Evaluate $\int \frac{(x^2-1)dx}{(x^4+3x^2+1)ta{n}^{-1}(x+\frac{1}{x})}$

Solution:

The given integral can be written as

$I=\int \frac{(1-\frac{1}{x^2})dx}{[(x+\frac{1}{x}{)}^2+1]ta{n}^{-1}(x+\frac{1}{x})}$

Let $(x+\frac{1}{x})=t$

Differentiating we get $(1-\frac{1}{x^2})dx=dt$

Hence $I=\int \frac{dt}{(t^2+1)ta{n}^{-1}t}$

Now make one more substitution $ta{n}^{–1}t=u$. Then $\frac{dt}{t^2+1}$and $I=\frac{du}{u}=ln|u|+C$

Returning to t, and then to x, we have

$I=ln|ta{n}^{-1}t|+C=ln|ta{n}^{-1}(x+\frac{1}{x})|+C$= Ans.

Example 8: Evaluate $\int \frac{co{s}^{3}x}{si{n}^{2}x+sinx}dx$

Solution:

$I=\int \frac{(1-si{n}^{2}x)cosx}{sinx(1+sinx)}dx=\int \frac{1-sinx}{sinx}cosxdx$

$Putsinx=t\Rightarrow cosxdx=dt$

$\Rightarrow I=\int \frac{1-t}{t}dt=ln|t|-t+C=ln|sinx|-sinx+C$= Ans.

Example 9: $\int \frac{dx}{sin(x+5)sin(x+8)}$

Solution :

Multiplying & dividing by sin(8 – 5)

$\Rightarrow \frac{1}{sin(8-5)}\int \frac{sin(8-5)}{sin(x+5)sin(x+8)}dx$

$\Rightarrow \frac{1}{sin3}\int \frac{sin((x+8)-(x+5))}{sin(x+5)sin(x+8}dx$

$\Rightarrow \frac{1}{sin3}\int \frac{sin(x+8)cos(x+5)-cos(x+8)sin(x+5)}{sin(x+5)sin(x+8)}dx$

$\Rightarrow \frac{1}{sin3}\int (cot(x+5)-cot(x+8))dx$

$\frac{1}{sin3}(ln(sin(x+5))-ln(sin(x+8)))+C$

7.0Sample Questions On Integration By Substitution 

1. What is the Formula for Integration by Substitution?

Ans: The formula is: $\int f(g(x)).g^{\prime }(x)dx=\int f(u)du$

Where u = g(x), and du = g'(x) dx.

2. How Do I Choose the Substitution in Integration by Substitution?

Ans: Look for a function inside another function (like $e^{x^2}$, $sin(3x)$, or $\frac{1}{x^2+1}$) and set the inside function as the substitution.