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JEE Mains Maths · Definite Integration

Evaluating by Substitution, Parts and Partial Fractions

Working out a definite integral directly: a substitution with its limits changed, integration by parts, partial fractions and inverse-tangent splits, finding the integrand before integrating, and bounding an integral that cannot be evaluated.

All Definite Integration notesJEE Mains Maths strategy

Why this matters

Forty-nine PYQs, a quarter of the chapter. Most need one well-chosen substitution; the rest integrate by parts, split into partial fractions, or first work out what the integrand is. Four ideas cover the page.

Concept 1 of 4: Substitution: tan x, tan(x/2) and rationalising

Choose a substitution that turns the integrand into a rational function or a standard form, and change the limits along with the variable, so there is nothing to substitute back. Even powers of sin⁡x\sin x and cos⁡x\cos x divided by cos⁡4x\cos^4x or similar become polynomials in t=tan⁡xt=\tan x. A denominator a+bsin⁡x+ccos⁡xa+b\sin x+c\cos x becomes rational with t=tan⁡x2t=\tan\frac x2. A difference of square roots is rationalised first.

Definition

  • Change the limits with the variable; never substitute back.
  • t=tan⁡xt=\tan x: dx=dt1+t2dx=\frac{dt}{1+t^2}, sec⁡2x=1+t2\sec^2x=1+t^2.
  • t=tan⁡x2t=\tan\frac x2: sin⁡x=2t1+t2\sin x=\frac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x=\frac{1-t^2}{1+t^2}, dx=2 dt1+t2dx=\frac{2\,dt}{1+t^2}.
  • 1a+b=a−ba−b\frac1{\sqrt a+\sqrt b}=\frac{\sqrt a-\sqrt b}{a-b}.

Half-angle substitution

t=tan⁡x2:sin⁡x=2t1+t2, cos⁡x=1−t21+t2t=\tan\frac x2:\quad \sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2}

Worked example

Evaluate ∫0π/2dx2+cos⁡x\int_0^{\pi/2}\frac{dx}{2+\cos x}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q66Moderate

Example 1 · Definite Integration · Evaluating by Substitution, Parts and Partial Fractions

The integral ∫0π213+2sin⁡x+cos⁡xdx\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x}dx is equal to:

The limits change too

With t=tan⁡xt=\tan x, the limit x=π2x=\frac\pi2 becomes t→∞t\to\infty, and with t=tan⁡x2t=\tan\frac x2 it becomes t=1t=1. Keeping the old limits is the most common slip.
Drill 22 more on substitution: tan x, tan(x/2) and rationalising

Concept 2 of 4: Parts, partial fractions and inverse-tangent splits

By parts moves a derivative from one factor to the other: ∫abuv′=[uv]ab−∫abu′v\int_a^b uv'=[uv]_a^b-\int_a^b u'v, with the boundary term worked out at the limits. A rational integrand splits into partial fractions. An inverse tangent of a quadratic often splits as a difference: cot⁡−1(1+x+x2)=tan⁡−1(x+1)−tan⁡−1x\cot^{-1}(1+x+x^2)=\tan^{-1}(x+1)-\tan^{-1}x, because (x+1)−x1+x(x+1)=11+x+x2\frac{(x+1)-x}{1+x(x+1)}=\frac1{1+x+x^2}.

Definition

  • ∫abuv′ dx=[uv]ab−∫abu′v dx\int_a^b uv'\,dx=[uv]_a^b-\int_a^b u'v\,dx.
  • Partial fractions before integrating a rational function.
  • tan⁡−1A−tan⁡−1B=tan⁡−1A−B1+AB\tan^{-1}A-\tan^{-1}B=\tan^{-1}\frac{A-B}{1+AB}: look for A−B=1A-B=1.
  • Inverse pair: ∫abf+∫f(a)f(b)f−1=bf(b)−af(a)\int_a^bf+\int_{f(a)}^{f(b)}f^{-1}=bf(b)-af(a) for increasing ff.

By parts

∫abu v′ dx=[uv]ab−∫abu′ v dx\int_a^b u\,v'\,dx=\big[uv\big]_a^b-\int_a^b u'\,v\,dx

Worked example

Evaluate ∫01xex dx\int_0^1xe^x\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q70Moderate

Example 2 · Definite Integration · Evaluating by Substitution, Parts and Partial Fractions

The integral ∫01cot⁡−1(1+x+x2)dx\int_{0}^{1} \cot^{- 1}\left( 1 + x +x^{2} \right)dx is equal to :

Work out the boundary term

[uv]ab[uv]_a^b is a number, not zero by default. It is zero only when uvuv vanishes at both limits; check before dropping it.
Drill 14 more on parts, partial fractions and inverse-tangent splits

Concept 3 of 4: Find the integrand first, then integrate

Many questions hide the integrand: a polynomial given through f(x2+1)f(x^2+1), a function fixed by a functional equation, or coefficients fixed by conditions. Find the function explicitly, check it against every condition, then integrate. Powers of sine and cosine are easier after reducing to multiple angles.

Definition

  • f(g(x))f(g(x)) given: put t=g(x)t=g(x) to recover f(t)f(t).
  • Functional equation in f(x)f(x) and f(1x)f\left(\frac1x\right): substitute again and solve the pair.
  • cos⁡2x=1+cos⁡2x2\cos^2x=\frac{1+\cos2x}2; cos⁡4x=38+12cos⁡2x+18cos⁡4x\cos^4x=\frac38+\frac12\cos2x+\frac18\cos4x.

Fourth power of cosine

cos⁡4x=38+12cos⁡2x+18cos⁡4x\cos^4x=\frac38+\frac12\cos2x+\frac18\cos4x

Worked example

f(x−1)=x2−2x+2f(x-1)=x^2-2x+2. Find ∫03f(x) dx\int_0^3f(x)\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q62Moderate

Example 3 · Definite Integration · Evaluating by Substitution, Parts and Partial Fractions

Let ff be a polynomial function such that f(x2+1)=x4+5x2+2f\left( x^{2}+ 1 \right)=x^{4}+ 5x^{2}+ 2, for all x∈Rx\in\mathbb{R}. Then ∫03f(x)dx\int_{0}^{3} f(x)dx is equal to

Check the function you found

A function recovered from one condition may fail another. Test it against every given value before integrating.
Drill 7 more on find the integrand first, then integrate

Concept 4 of 4: Bounding an integral without evaluating it

If m≤f(x)≤Mm\le f(x)\le M on [a,b][a,b], then m(b−a)≤∫abf≤M(b−a)m(b-a)\le\int_a^bf\le M(b-a). A monotonic integrand takes its extreme values at the ends, which gives the tightest such bounds. Split the interval where the bounds on ff change.

Definition

  • m≤f≤Mm\le f\le M on [a,b]⇒m(b−a)≤∫abf≤M(b−a)[a,b]\Rightarrow m(b-a)\le\int_a^bf\le M(b-a).
  • ff decreasing: (b−a)f(b)≤∫abf≤(b−a)f(a)(b-a)f(b)\le\int_a^bf\le(b-a)f(a).
  • Different bounds on different parts: add the parts.

Bounds from the extreme values

m(b−a)≤∫abf(x) dx≤M(b−a)m(b-a)\le\int_a^b f(x)\,dx\le M(b-a)

Worked example

Between which numbers does ∫01ex2dx\int_0^1e^{x^2}dx lie?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q68Moderate

Example 4 · Definite Integration · Evaluating by Substitution, Parts and Partial Fractions

Let g(x)=∫0xf(t)dtg(x) =\int_{0}^{x} f(t)dt, where ff is continuous function in [0,3]\lbrack 0,3\rbrack such that 13≤f(t)≤1\frac{1}{3}\leq f(t) \leq 1 for all t∈[0,1]t \in \lbrack 0,1\rbrack and 0≤f(t)≤120 \leq f(t) \leq\frac{1}{2} for all t∈(1,3]t \in (1,3\rbrack. The largest possible interval in which g(3)g(3) lies is:

Check the bound against the options

Work out the two bounds as decimals and compare with each option. If the bounds fit no option, recheck the monotonicity before trusting either.
Drill 2 more on bounding an integral without evaluating it

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Watch out for (4)

Test yourself on Definite Integration

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