What is the Weierstrass or universal substitution?

The substitution t = tan(x/2) converts any rational function of sin x and cos x into a rational function of t. Under this substitution sin x = 2t/(1+t^2), cos x = (1-t^2)/(1+t^2), dx = 2 dt/(1+t^2). It always works but often produces cumbersome calculations.

Why does the integral of f prime of x over f of x equal ln|f(x)| + C?

Put f(x) = t so f prime of x dx = dt. The integral becomes the integral of dt over t which equals ln|t| + C = ln|f(x)| + C. Any time the numerator is exactly the derivative of the denominator, the answer is the logarithm of the denominator.

Methods of Integration

Q: When should I use integration by substitution vs. by parts?

Use substitution when the integrand contains a function g(x) and its derivative up to a constant, or when a substitution eliminates a radical or composite. Use integration by parts when the integrand is a product of two unrelated functions like x sin x or x ln x.

Q: What is a derived substitution or algebraic twin?

Integrals like the integral of 1/(x^4 + 1) do not yield to a single substitution. The trick is to split them into a sum or difference of two related integrals then use t = x - 1/x on one and t = x + 1/x on the other.

Q: Do I need to memorise all the standard substitutions?

Yes. The main radical-based substitutions x = a sin theta, a tan theta, a sec theta, the sqrt((x-a)(b-x)) trick, and t = tan(x/2) appear in nearly every JEE calculus paper. Being able to identify the right substitution in under 10 seconds is a major advantage.

Q: Can I always change back to x at the end of substitution?

Yes, and you must. Since t or theta was introduced only as a tool, the substitution t = g(x) is invertible on the domain in question, and the final answer must express the antiderivative as a function of x.

MathsIntegralsFor JEE aspirants

The three main methods of integration used in JEE Maths are (1) integration by substitution, (2) integration by parts, and (3) integration by partial fractions. This concept focuses on the substitution method - by far the most frequently used technique. Given an integral $\int f (g (x)) g^{'} (x) d x$ , the substitution $t = g (x)$ converts it into $\int f (t) d t$ , which is often a standard integral. Substitution is classified into direct, standard-trig, indirect, and derived (algebraic/trigonometric twins). The other two methods - parts and partial fractions - are covered in separate concepts.

Key Formulas - Quick Reference

$\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + C$
$\int \frac{f ^{'} ( x )}{f ( x )} d x = 2 f (x) + C$
$\int [f (x)]^{n} f^{'} (x) d x = \frac{[ f ( x ) ] ^{n + 1}}{n + 1} + C, n \neq = - 1$
For $a^{2} - x^{2}$ : put $x = a sin θ$ (or $a cos θ$ )
For $x^{2} + a^{2}$ : put $x = a tan θ$ (or $a cot θ$ )
For $x^{2} - a^{2}$ : put $x = a sec θ$ (or $a csc θ$ )
For $(x - a) (b - x)$ : put $x = a cos^{2} θ + b sin^{2} θ$
Weierstrass (for trig rational): put $t = tan (x /2)$ ; then $sin x = \frac{2 t}{1 + t ^{2}}, cos x = \frac{1 - t ^{2}}{1 + t ^{2}}, d x = \frac{2 d t}{1 + t ^{2}}$

1. The Three Methods at a Glance

If the integrand is not itself a standard derivative, we need a technique to reduce it to a known form. The three principal methods are:

Integration by substitution (change of variable) - covered in this concept.
Integration by parts - for products of two functions (covered separately).
Integration by partial fractions - for rational functions $P (x) / Q (x)$ (covered separately).

2. Integration by Substitution

The idea: choose a new variable $t = g (x)$ so that the integrand simplifies. Formally, if $\int f (x) d x = F (x) + C$ , then

$\int f (g (x)) g^{'} (x) d x = F (g (x)) + C .$

In practice, this is executed by putting $g (x) = t$ , so that $g^{'} (x) d x = d t$ , converting the integral into $\int f (t) d t$ .

2.1 Direct Substitution

When the integral is of the form $\int f (g (x)) g^{'} (x) d x$ , put $g (x) = t$ . Two special cases arise so often that they deserve to be memorised as formulae:

(i) $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + C .$ (Put $f (x) = t \Rightarrow f^{'} (x) d x = d t$ .)

(ii) $\int \frac{f ^{'} ( x )}{f ( x )} d x = 2 f (x) + C .$ (Put $f (x) = t$ .)

Solved Example 1

Evaluate

\int \frac{sin ( ln x )}{x} d x .

Solution:

Put $ln x = t$ . Then $\frac{1}{x} d x = d t$ .

$I = \int sin t d t = - cos t + C = - cos (ln x) + C .$

Solved Example 2

Evaluate

\int \frac{x ^{3} tan ^{- 1} ( x ^{4} )}{1 + x ^{8}} d x .

Solution:

Put $tan^{- 1} (x^{4}) = t$ . Differentiating:

$\frac{4 x ^{3}}{1 + x ^{8}} d x = d t \Rightarrow \frac{x ^{3}}{1 + x ^{8}} d x = \frac{1}{4} d t .$

Therefore

$I = \frac{1}{4} \int t d t = \frac{1}{4} \cdot \frac{t ^{2}}{2} + C = \frac{1}{8} [tan^{- 1} (x^{4})]^{2} + C .$

Solved Example 3

Evaluate

\int (2 x^{2} + 1) sin^{2} (2 x^{3} + 3 x) d x .

Solution:

Let $z = 2 x^{3} + 3 x$ . Then

$d z = (6 x^{2} + 3) d x = 3 (2 x^{2} + 1) d x \Rightarrow (2 x^{2} + 1) d x = \frac{1}{3} d z .$

Substituting and using $sin^{2} z = \frac{1 - cos 2 z}{2}$ :

$I = \frac{1}{3} \int sin^{2} z d z = \frac{1}{6} \int (1 - cos 2 z) d z = \frac{1}{6} [z - \frac{sin 2 z}{2}] + C .$

Substituting back $z = 2 x^{3} + 3 x$ :

$I = \frac{1}{6} [2 x^{3} + 3 x - \frac{sin ( 4 x ^{3} + 6 x )}{2}] + C .$

2.2 Standard Trigonometric Substitutions

Certain algebraic radicals become tractable under trigonometric substitutions. Memorise this table:

Term in integrand	Substitution
$x^{2} + a^{2}$ or $x^{2} + a^{2}$	$x = a tan θ$ (or $a cot θ$ )
$x^{2} - a^{2}$ or $x^{2} - a^{2}$	$x = a sec θ$ (or $a csc θ$ )
$a^{2} - x^{2}$ or $a^{2} - x^{2}$	$x = a sin θ$ (or $a cos θ$ )
Both $a + x$ and $a - x$ present	$x = a cos θ$ (or $a cos 2 θ$ )
$(x - a) (b - x)$	$x = a cos^{2} θ + b sin^{2} θ$
$(x^{2} + a^{2} \pm x)^{n}$ or $(x \pm x^{2} - a^{2})^{n}$	Put the entire bracket $= t$
$\frac{1}{( x + a ) ^{n_{1}} ( x + b ) ^{n_{2}}}, n_{1}, n_{2} > 1$	Put $(x + a) = t (x + b)$

Figure 1: Right-triangle picture behind the substitution

x = a tan θ

, giving

x^{2} + a^{2} = a sec θ

and

d x = a sec^{2} θ d θ

Solved Example 4

Evaluate

\int \frac{1 - x}{1 + x} d x .

Solution:

Rationalise by multiplying inside the radical by $\frac{1 - x}{1 - x}$ :

$\frac{1 - x}{1 + x} = \frac{1 - x}{1 - x} .$

$I = \int \frac{d x}{1 - x} - \int \frac{x}{1 - x} d x .$

The first integral evaluates to $- 2 1 - x$ . For the second, put $x = sin^{2} ϕ$ , so $d x = 2 sin ϕ cos ϕ d ϕ = sin 2 ϕ d ϕ$ and $1 - x = cos ϕ, x = sin ϕ$ :

$\int \frac{sin ϕ}{cos ϕ} \cdot sin 2 ϕ d ϕ = \int \frac{sin ϕ}{cos ϕ} \cdot 2 sin ϕ cos ϕ d ϕ = 2 \int sin^{2} ϕ d ϕ = ϕ - \frac{sin 2 ϕ}{2} .$

Since $sin ϕ = x$ , we get $ϕ = sin^{- 1} x$ and $\frac{sin 2 ϕ}{2} = sin ϕ cos ϕ = x 1 - x$ . Therefore

$I = - 2 1 - x - sin^{- 1} x + x 1 - x + C .$

Solved Example 5

Evaluate

\int \frac{d x}{( x + 1 ) ^{6/5} ( x - 3 ) ^{4/5}} .

Solution:

Rewrite by pulling out a factor $(x + 1)^{2}$ from the denominator:

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

Put $\frac{x - 3}{x + 1} = t$ . Differentiating:

$d t = \frac{( x + 1 ) - ( x - 3 )}{( x + 1 ) ^{2}} d x = \frac{4}{( x + 1 ) ^{2}} d x \Rightarrow \frac{d x}{( x + 1 ) ^{2}} = \frac{d t}{4} .$

Hence

$I = \frac{1}{4} \int \frac{d t}{t ^{4/5}} = \frac{1}{4} \cdot \frac{t ^{1/5}}{1/5} + C = \frac{5}{4} (\frac{x - 3}{x + 1})^{1/5} + C .$

2.3 Indirect Substitution

If the integrand is of the form $f (x) g (x)$ , where $g (x)$ is a function of the integral of $f (x)$ , then put $\int f (x) d x = t$ .

Solved Example 6

Evaluate

\int \frac{x}{x ^{3} + a ^{3}} d x .

Solution:

The antiderivative of $x = x^{1/2}$ is $\frac{x ^{3/2}}{3/2}$ , so put $x^{3/2} = t$ . Then $\frac{3}{2} x^{1/2} d x = d t$ , i.e. $x d x = \frac{2}{3} d t$ .

Also $x^{3} = t^{2}$ , so $x^{3} + a^{3} = t^{2} + a^{3}$ . Therefore

$I = \frac{2}{3} \int \frac{d t}{t ^{2} + a ^{3}} = \frac{2}{3} ln x^{3/2} + x^{3} + a^{3} + C,$

using the standard formula $\int \frac{d t}{t ^{2} + k ^{2}} = ln t + t^{2} + k^{2} + C$ .

2.4 Derived Substitutions - Algebraic and Trigonometric Twins

Sometimes an integral splits naturally into two related integrals that can each be tackled by a clever substitution. Common examples:

Algebraic twins: $\int \frac{2 x ^{2}}{x ^{4} + 1} d x, \int \frac{2}{x ^{4} + 1} d x, \int \frac{2 x ^{2}}{x ^{4} + k x ^{2} + 1} d x .$

Trigonometric twins: $\int tan x d x, \int cot x d x, \int \frac{d x}{sin ^{4} x + cos ^{4} x}, \int \frac{d x}{sin ^{6} x + cos ^{6} x} .$

The trick: divide numerator and denominator by $x^{2}$ (algebraic) or by $cos^{2} x$ / $sin^{2} x$ (trigonometric), then use the substitutions $t = x - 1/ x$ or $t = x + 1/ x$ .

Solved Example 7

Evaluate

\int \frac{5}{1 + x ^{4}} d x .

Solution:

Write $\frac{5}{1 + x ^{4}} = \frac{5}{2} [\frac{( x ^{2} + 1 ) + ( 1 - x ^{2} )}{x ^{4} + 1}] \cdot \frac{2}{2} = \frac{5}{2} [\frac{x ^{2} + 1}{x ^{4} + 1} + \frac{1 - x ^{2}}{x ^{4} + 1}] .$

Divide numerator and denominator of each by $x^{2}$ :

$\frac{x ^{2} + 1}{x ^{4} + 1} = \frac{1 + 1/ x ^{2}}{x ^{2} + 1/ x ^{2}}, \frac{1 - x ^{2}}{x ^{4} + 1} = - \frac{1 - 1/ x ^{2}}{x ^{2} + 1/ x ^{2}} .$

For $I_{1} = \int \frac{1 + 1/ x ^{2}}{x ^{2} + 1/ x ^{2}} d x$ : put $x - \frac{1}{x} = t$ , so $(1 + \frac{1}{x ^{2}}) d x = d t$ , and $x^{2} + \frac{1}{x ^{2}} = t^{2} + 2$ :

$I_{1} = \int \frac{d t}{t ^{2} + ( 2 ) ^{2}} = \frac{1}{2} tan^{- 1} (\frac{x - 1/ x}{2}) .$

For $I_{2} = \int \frac{1 - 1/ x ^{2}}{x ^{2} + 1/ x ^{2}} d x$ : put $x + \frac{1}{x} = t$ , so $(1 - \frac{1}{x ^{2}}) d x = d t$ , and $x^{2} + \frac{1}{x ^{2}} = t^{2} - 2$ :

$I_{2} = \int \frac{d t}{t ^{2} - ( 2 ) ^{2}} = \frac{1}{2 2} ln \frac{x + 1/ x - 2}{x + 1/ x + 2} .$

Combining (note the sign of $I_{2}$ flipped when we split):

$I = \frac{5}{2} [\frac{1}{2} tan^{- 1} (\frac{x - 1/ x}{2}) + \frac{1}{2 2} ln \frac{x + 1/ x - 2}{x + 1/ x + 2}] + C .$

Solved Example 8

Evaluate

\int tan x d x .

Solution:

Put $tan x = t^{2}$ , so $sec^{2} x d x = 2 t d t$ , i.e. $d x = \frac{2 t d t}{1 + t ^{4}}$ . Then

$I = \int t \cdot \frac{2 t d t}{1 + t ^{4}} = \int \frac{2 t ^{2} d t}{1 + t ^{4}} = \int \frac{t ^{2} + 1}{t ^{4} + 1} d t + \int \frac{t ^{2} - 1}{t ^{4} + 1} d t .$

Each integral is exactly the algebraic-twins form of Example 7. Applying the same substitutions ( $t - 1/ t = u$ for the first, $t + 1/ t = u$ for the second) gives

$I = \frac{1}{2} tan^{- 1} (\frac{t ^{2} - 1}{t 2}) + \frac{1}{2 2} ln \frac{t ^{2} + 1 - t 2}{t ^{2} + 1 + t 2} + C,$

with $t = tan x$ .

Common Mistakes to Avoid

Watch out

Forgetting to change $d x$ when substituting. If $t = g (x)$ , then $d t = g^{'} (x) d x$ , so $d x = d t / g^{'} (x)$ . The substitution is incomplete until both the integrand and the differential are rewritten in terms of $t$ .
Not substituting back to $x$ at the end. The final answer of an indefinite integral must be expressed in the original variable $x$ , not the substituted variable $t$ or $θ$ .
Choosing a substitution that doesn't simplify. A good substitution eliminates a radical or reduces a composite function to a single symbol. If your integral in $t$ looks harder than the one in $x$ , try a different substitution.
Confusing $\int \frac{f ^{'} ( x )}{f ( x )} d x$ with $\int \frac{1}{f ( x )} d x$ . The first is $ln ∣ f (x) ∣ + C$ ; the second has no general formula and usually requires partial fractions or another technique.
Using $x = a sin θ$ when the integrand has $x^{2} + a^{2}$ . $a^{2} - x^{2}$ needs $sin θ$ ; $x^{2} + a^{2}$ needs $tan θ$ . Mixing them produces imaginary intermediate expressions.

Frequently Asked Questions

Q1. When should I use integration by substitution vs. by parts?

Use substitution when the integrand contains a function $g (x)$ and its derivative $g^{'} (x)$ (up to a constant factor), or when a substitution eliminates a radical or a composite. Use integration by parts when the integrand is a product of two unrelated functions - typically an algebraic times a transcendental function like $x sin x$ or $x ln x$ .

Q2. How do I know which substitution to try for a given integral?

Look for a function whose derivative already appears in the integrand (direct substitution). If the integrand contains $a^{2} \pm x^{2}$ or $x^{2} - a^{2}$ , use the standard trigonometric substitutions ( $x = a sin θ$ , $a tan θ$ , or $a sec θ$ ). If it's a rational function of $sin x$ and $cos x$ , try $t = tan (x /2)$ .

Q3. What is the Weierstrass (universal) substitution?

The substitution $t = tan (x /2)$ converts any rational function of $sin x$ and $cos x$ into a rational function of $t$ , which can then be integrated by partial fractions. Under this substitution, $sin x = 2 t / (1 + t^{2}), cos x = (1 - t^{2}) / (1 + t^{2}), d x = 2 d t / (1 + t^{2})$ . It always works but often produces complicated calculations - simpler substitutions should be tried first.

Q4. What does "derived substitution" or "algebraic twin" mean?

Integrals like $\int d x / (x^{4} + 1)$ don't yield to a single substitution. The trick is to split them into a sum or difference of two related integrals, then use $t = x - 1/ x$ on one and $t = x + 1/ x$ on the other. The two "twin" integrals are companions - the algebra unifies them.

Q5. Do I need to memorise all the standard substitutions?

Yes. The five main radical-based substitutions ( $x = a sin θ, a tan θ, a sec θ$ , the $(x - a) (b - x)$ trick, and $t = tan (x /2)$ ) appear in nearly every JEE calculus paper. Time saved on recognition compounds - being able to identify the right substitution in under 10 seconds is a major advantage.

Q6. Why does $\int f^{'} (x) / f (x) d x = ln ∣ f (x) ∣ + C$ ?

Put $f (x) = t$ , so $f^{'} (x) d x = d t$ . The integral becomes $\int d t / t = ln ∣ t ∣ + C = ln ∣ f (x) ∣ + C$ . This is one of the most useful shortcuts in integration - any time the numerator is exactly the derivative of the denominator, the answer is the logarithm of the denominator.

Q7. Can I always change back to $x$ at the end?

Yes, and you must. Since $t$ (or $θ$ ) was introduced only as a tool, the substitution $t = g (x)$ is invertible on the domain in question, and the final answer must express the antiderivative as a function of $x$ . If the back-substitution looks ugly, that's usually a sign the answer is genuinely complicated - not that the method is wrong.

Previous year questions on Methods of Integration

13 questions from past papers, each with a step-by-step solution.

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1. The Three Methods at a Glance

2. Integration by Substitution

2.1 Direct Substitution

2.2 Standard Trigonometric Substitutions

2.3 Indirect Substitution

2.4 Derived Substitutions - Algebraic and Trigonometric Twins

Common Mistakes to Avoid

Frequently Asked Questions

Q1. When should I use integration by substitution vs. by parts?

Q2. How do I know which substitution to try for a given integral?

Q3. What is the Weierstrass (universal) substitution?

Q4. What does "derived substitution" or "algebraic twin" mean?

Q5. Do I need to memorise all the standard substitutions?

Q6. Why does ∫f′(x)/f(x)dx=ln∣f(x)∣+C?

Q7. Can I always change back to x at the end?

Previous year questions on Methods of Integration

Ready to master Integrals?

Q6. Why does $\int f^{'} (x) / f (x) d x = ln ∣ f (x) ∣ + C$ ?

Q7. Can I always change back to $x$ at the end?