Indirect Substitution in Integral

Indirect Substitution in Integral

Sometimes when you’re solving an integral, it feels a bit like trying to open a tightly jammed jar—no matter how much you twist, it just doesn’t budge. And then someone lightly taps the lid, and suddenly it opens in one go. Indirect substitution works exactly like that. Instead of forcing a direct substitution, you make a clever detour that simplifies the entire expression. This technique is especially powerful for complex algebraic and trigonometric integrals that resist straightforward methods.

What Is Indirect Substitution?

Indirect substitution is a strategic integration technique where, instead of substituting the integrand directly, we transform it into a related form that makes substitution easier. This creates a bridge between the original integral and a more manageable one—especially useful in competitive exams like JEE, CUET, and board-level calculus.

This method shines when:

• The integrand has reciprocal expressions such as $x + \frac{1}{x}$

• The expression can be rewritten as sum/difference of similar structures

• The integral contains algebraic or trigonometric twins

• The integrand is a product of two functions, where one becomes the derivative of the other after transformation

Why Use Indirect Substitution?

Indirect substitution is chosen when:

• Direct substitution doesn’t simplify the integrand

• The expression becomes friendlier only after rearrangement

• The integral resembles a known standard form after transformation

• Traditional techniques like IBP or partial fractions fail

It’s a quiet trick in the calculus toolkit, but extremely powerful.

Step-by-Step Approach to Indirect Substitution

In this section, a concise step-by-step approach to indirect substitution is outlined, guiding you on how to convert complex integrals into simpler ones through appropriate variable changes, followed by rewriting and evaluating the integral in the new variable before reversing the substitution.

1. Identify patterns like $x + \frac{1}{x}$ or $x - \frac{1}{x}$

These are perfect candidates for indirect substitution.

2. Rewrite the integrand in terms of reciprocal expressions

This often reveals hidden structure.

3. Introduce a substitution that simplifies the entire form

Example: $t = x + \frac{1}{x}$ or $t = \tan x$

4. Transform the differential using algebraic manipulation

This is the heart of indirect substitution.

5. Integrate the simpler transformed integral

Most of the time, you get a rational function or a standard integral.

Illustration 1: Evaluate

$\int \frac{3x^4 + 4x^3}{(x^4 + x + 1)^2} , dx$

Rewrite:
$I = \int \frac{3x^4 + 4x^3}{(x^4 + x + 1)^2} dx$

Divide numerator and denominator by $x^8$ to reveal the structure:
$I = \int \frac{\frac{3}{x^4} + \frac{4}{x^5}}{\left(1 + \frac{1}{x^3} + \frac{1}{x^4}\right)^2} dx$

Let
$1 + \frac{1}{x^3} + \frac{1}{x^4} = t$

Then
$\left(-\frac{3}{x^4} - \frac{4}{x^5}\right) dx = dt$

So
$I = -\int \frac{dt}{t^2} = \frac{1}{t} + C$

Back-substitute:
$I = \frac{x^4}{x^4 + x + 1} + C$

Illustration 2: Evaluate

$\int (x^{3m} + x^{2m} + x^m)(2x^{2m} + 3x^m + 6)^{1/m} dx,; x>0$

Rewrite by dividing one $x$:
$I = \int (x^{3m-1} + x^{2m-1} + x^{m-1})(2x^{3m} + 3x^{2m} + 6x^m)^{1/m} dx$

Let
$2x^{3m} + 3x^{2m} + 6x^m = t$

Then
$dt = 6m(x^{3m-1} + x^{2m-1} + x^{m-1}) dx$

Thus,
$I = \frac{1}{6m} \int t^{1/m} dt = \frac{1}{6(m+1)} t^{\frac{m+1}{m}} + C$

Back-substituting gives the final expression.

Algebraic Twins in Indirect Substitution

Algebraic twins are pairs of integrals that look different but share a deep internal structure. Solving one automatically helps solve the other.

Examples:
$\int \frac{2x^2}{x^4 + 1} dx$
can be split into two simpler integrals:
$\int \frac{x^2 + 1}{x^4 + 1} dx + \int \frac{x^2 - 1}{x^4 + 1} dx$

Standard Patterns Useful in Indirect Substitution

Standard patterns useful in indirect substitution are commonly recurring forms or structures in integrals that allow simplification through substitution techniques. Recognizing these patterns helps transform complex integrals into manageable forms by choosing appropriate substitutions based on algebraic or trigonometric identities.

1. Expressions with $x + \frac{1}{x}$

Form:
$\int f\left(x + \frac{1}{x}\right)\left(1 - \frac{1}{x^2}\right) dx$

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

Then
$\left(1 - \frac{1}{x^2}\right) dx = dt$

2. Expressions with $x - \frac{1}{x}$

Form:
$\int f\left(x - \frac{1}{x}\right)\left(1 + \frac{1}{x^2}\right) dx$

Let
$x - \frac{1}{x} = t$

Then
$\left(1 + \frac{1}{x^2}\right) dx = dt$

3. Rational Integrals with Quartics

Forms like
$\int \frac{x^2 + 1}{x^4 + kx^2 + 1} dx$
Divide numerator and denominator by $x^2$ to simplify structure.

Solved Examples Based on Indirect Substitution in Integral

Example 1: Evaluate $\int \frac{\left(x^2-1\right) d x}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}$

1) $-\ln \left|\tan ^{-1}\left(x-\frac{1}{x}\right)\right|+c$
2) $\ln \left|\tan ^{-1}\left(x-\frac{1}{x}\right)\right|+c$
3) $\left.\ln \tan ^{-1}\left(x+\frac{1}{x}\right) \right\rvert\,+c$
4) $-\ln \tan ^{-1}\left(x+\frac{1}{x}\right)+c$

Solution

Put $\left(x+\frac{1}{x}\right)=t$

Integral can be written as

$\int \frac{\left(1-\frac{1}{x^2}\right) d x}{\left[\left(x+\frac{1}{x}\right)^2+1\right] \tan ^{-1}\left(x+\frac{1}{x}\right)}$

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

Differentiating we get $\left(1-\frac{1}{x^2}\right) d x=d t$

Hence, $I=\int \frac{d t}{\left(t^2+1\right) \tan ^{-1} t}$

Now make one more substitution tan-1t = u. Then

$\frac{d t}{t^2+1}=d u$ and $\mathrm{I}=\int \frac{\mathrm{du}}{11}=\ln |u|+c$

$I=\ln \left|\tan ^{-1} t\right|+c=\ln \left|\tan ^{-1}\left(x+\frac{1}{x}\right)\right|+c$

Hence, the answer is the option 3.

Example 2: The value of the integral $\int\left(x+\frac{1}{x}\right)^{n+5}\left(\frac{x^2-1}{x^2}\right) d x$ is equal to

11) $\frac{\left(x+\frac{1}{x}\right)^{n+6}}{n+6}+c$
2) $\left(\frac{x^2+1}{x^2}\right)^{n+6}(n+6)+c$
3) $\left(\frac{x}{x^2+1}\right)^{n+6}(n+6)+c$

4) None of these

Solution

We put $\left(x+\frac{1}{x}\right)=t$

$I=\int p^{n+5} d p$ If $x+\frac{1}{x}=p$ then, $\left(1-\frac{1}{x^2}\right) d x=d p$

$\begin{aligned} & \quad \therefore I=\int\left(x+\frac{1}{x}\right)^{n+3} \quad\left(\frac{x^2-1}{x^2}\right) d x=\int p^{n+5} d p=\frac{p^{n+6}}{n+6}+c= \\ & \frac{\left(x+\frac{1}{x}\right)^{n+6}}{n+6}+c\end{aligned}$

Hence, the answer is the option 1.

Example 3: The integral $\int \frac{d x}{(x+1)^{\frac{3}{4}}(x-2)^{\frac{5}{4}}}$ is equal to:

1) $4\left(\frac{x+1}{x-2}\right)^t+C$
2) $4\left(\frac{x-2}{x+1}\right)^t+C$
3) $-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}}+C$
4) $-\frac{4}{3}\left(\frac{x-2}{x+1}\right)^{\frac{1}{4}}+C$

Solution

Integration by substitution -

$I=\frac{d x}{(x+1)^{\frac{3}{4}}(x-2)^{\frac{5}{4}}}$

$I=\int \frac{d x}{(x+1)^{\frac{3}{4}}(x-2)^{\frac{5}{4}}} \frac{(x-2)^{\frac{3}{4}}}{(x-2)^{\frac{3}{4}}}$

$=\int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{\frac{3}{4}}(x-2)^2}$

Let $\frac{x+1}{x-2}=t$

Differentiating 1) on both sides

$\frac{(x-2)-(x+1)}{(x-2)^2} d x=d t$

$\Rightarrow \frac{d x}{(x-2)^2}=-\frac{d t}{3}$

Thus I= $\int \frac{-d t}{3 t^{\frac{3}{4}}}$

$=\frac{-1}{3} \times \frac{1}{\frac{1}{4}} t^{\frac{1}{4}}+c$

$=-\frac{4}{3} t^{\frac{1}{4}}+c$

$I=-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}}+c$

Hence, the answer is the option 3.

Example 4: For x>0, let $f(x)=\int_1^x \frac{\log t}{1+t} d t$. Then $f(x)+f\left(\frac{1}{x}\right)$ is equalto

1) $\frac{1}{4}(\log x)^2$
2) $\frac{1}{2}(\log x)^2$
3) $\log x$
4) $\frac{1}{4} \log x^2$

Solution
$
\begin{aligned}
& f(x)=\int_1^x \frac{\log t}{1+t} d x \\
& f\left(\frac{1}{x}\right)=\int_1^{\frac{1}{2}} \frac{\log t}{1+t} d t
\end{aligned}
$
Put t= (substitution)

$
\begin{aligned}
& f\left(\frac{1}{x}\right)=\int_1^1 \frac{\log t}{1+t} d t=\int_1^2\left(\frac{-\log z}{1+\frac{1}{z}}\right) \times \frac{-1}{z^2} d z \\
& =\int_1^z\left(\frac{+\log z}{1+z}\right) \times \frac{1}{z} d z
\end{aligned}
$

Substituting z = x in (i) we get f(x).

Thus,

$\begin{aligned} & f(x)+f\left(\frac{1}{x}\right)=\int_1^x \frac{\log t}{1+t}\left(1+\frac{1}{t}\right) d t \\ & =\int_1^x \frac{\log t}{t} d t \\ & =\frac{(\log x)^2}{2}-0\end{aligned}$

Hence, the answer is the option 2.

Example 5: If $\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t=\frac{1}{2}(g(t))^2+C$ where C is a constant, then g(2) is equal to :

1) $2 \log (2+\sqrt{5})$
2) $\log (2+\sqrt{5})$
3) $\frac{1}{\sqrt{5}} \log (2+\sqrt{5})$
4) $\left.2^{\frac{1}{\log }(2+\sqrt{5}}\right)$

Solution

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity give any one of the standard formulas.

- wherein

Since $\int f(x) d x=\int f(t) d t=\int f(\theta) d \theta$ all variables must be converted into a single variable,$(\operatorname{tor} \theta)$

$\begin{aligned} & \int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t=\frac{1}{2}(g(t))^2+C \\ & L H S=\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t \\ & \frac{1}{t+\sqrt{1+t^2}} \times\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t=d m \\ & \frac{d t}{\sqrt{1+t^2}}=d m \\ & \text { we get } \int \log \left(t+\sqrt{1+t^2}\right)=m \\ & \text { LHS }=\frac{1}{2}\left[\log \left(1+\frac{m^2}{2}+C\right.\right. \\ & \left.\text { So, } g(t)=\log \left(1+\sqrt{1+t^2}\right)\right]^2+C \\ & \text { Put } t=2, \text { we get } g(2)=\log (2+\sqrt{5})\end{aligned}$

Hence, the answer is the option 2.

List of Topics Related to the Indirect Substitution in Integral

This section covers key topics related to indirect substitution in integrals, highlighting techniques that transform complex integrals into simpler forms through strategic variable changes. It connects foundational concepts like definite and indefinite integrals with substitution methods, enabling easier evaluation of integrals involving composite and irrational expressions. This approach deepens understanding and enhances skills in solving a variety of integral problems effectively.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by parts

Application of Inequality in Definite Integration

NCERT Resources

This section compiles comprehensive NCERT resources for Chapter 7 – Integrals, providing everything needed for thorough preparation. It includes detailed notes, step-by-step solutions, and challenging exemplar problems designed to clarify integral concepts and build strong problem-solving skills. These materials are ideal for both board exam readiness and competitive exam preparation, helping students gain confidence and mastery in integration techniques.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Indirect Substitution in Integral

This section provides a carefully curated collection of practice questions focused on indirect substitution in integrals. These exercises are designed to help you progressively strengthen your problem-solving abilities by applying substitution techniques to a variety of integral forms, including algebraic and trigonometric expressions. Regular practice will build your confidence and mastery in approaching integrals using indirect substitution effectively.

Indirect Substitution In Integral- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: