PYQ Vault

JEE Mains Maths · Indefinite Integration

Trigonometric Integrals

Integrals of powers and combinations of sin x and cos x, turned into algebra by putting t equal to tan x, cot x, sin x, cos x or sin x ± cos x.

All Indefinite Integration notesJEE Mains Maths strategy

Why this matters

Nine PYQs, seven of them multiple choice, and one from 2026. Four divide through by a power of cos x or sin x so that only tan x or cot x is left; five substitute sin x, cos x or sin x ± cos x, often after a double-angle identity. Two ideas cover the page.

Concept 1 of 2: Everything in tan x or cot x

When the powers of sin⁡x\sin x and cos⁡x\cos x add to a negative even number, divide through by a power of cos⁡x\cos x. What is left is a function of tan⁡x\tan x times sec⁡2x\sec^2x, and sec⁡2x dx\sec^2x\,dx is d(tan⁡x)d(\tan x). The same move works for asin⁡2x+bcos⁡2xa\sin^2x+b\cos^2x in a denominator. Use cot⁡x\cot x instead when the larger power is on sin⁡x\sin x.

Definition

  • sin⁡mxcos⁡nx\sin^mx\cos^nx with m+nm+n a negative even integer: write it through tan⁡x\tan x and sec⁡2x\sec^2x, or cot⁡x\cot x and csc⁡2x\csc^2x.
  • sec⁡2x=1+tan⁡2x\sec^2x=1+\tan^2x and csc⁡2x=1+cot⁡2x\csc^2x=1+\cot^2x turn the leftover factors into powers of tt.
  • 1asin⁡2x+bcos⁡2x\frac{1}{a\sin^2x+b\cos^2x}: divide top and bottom by cos⁡2x\cos^2x and put t=tan⁡xt=\tan x.
  • With t=cot⁡xt=\cot x, dt=−csc⁡2x dxdt=-\csc^2x\,dx.

Tan substitution

∫dxa2sin⁡2x+b2cos⁡2x=1abtan⁡−1(atan⁡xb)+C\int\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac1{ab}\tan^{-1}\left(\frac{a\tan x}{b}\right)+C

Worked example

Find ∫dxsin⁡xcos⁡3x\int\frac{dx}{\sin x\cos^3x}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q73Moderate

Example 1 · Indefinite Integration · Trigonometric Integrals

If ∫ (sin⁡x)−112(cos⁡x)−52dx=\int_{}^{}\ (\sin x)^{\frac{- 11}{2}}(\cos x)^{\frac{- 5}{2}}dx = −p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C- \frac{p_{1}}{q_{1}}(\cot x)^{\frac{9}{2}} - \frac{p_{2}}{q_{2}}(\cot x)^{\frac{5}{2}} - \frac{p_{3}}{q_{3}}(\cot x)^{\frac{1}{2}} + \frac{p_{4}}{q_{4}}(\cot x)^{\frac{- 3}{2}} + C, where pip_{i} and qiq_{i} are positive integers with gcd⁡(pi,qi)=1\gcd\left( p_{i},q_{i} \right) = 1 for i=1,2,3,4i = 1,2,3,4 and C is the constant of integration, then 15p1p2p3p4q1q2q3q4\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}} is equal to ____\_\_\_\_ .

Cot brings a minus sign

With t=cot⁡xt=\cot x, dt=−csc⁡2x dxdt=-\csc^2x\,dx. Every term of the answer changes sign; dropping the minus gives each coefficient the wrong sign.
Drill 3 more on everything in tan x or cot x

Concept 2 of 2: Substituting sin x, cos x or sin x ± cos x

If the integrand has a lone cos⁡x dx\cos x\,dx, put t=sin⁡xt=\sin x; if it has a lone sin⁡x dx\sin x\,dx, put t=cos⁡xt=\cos x. If the top is cos⁡x±sin⁡x\cos x\pm\sin x, it is the derivative of sin⁡x∓cos⁡x\sin x\mp\cos x, and sin⁡2x\sin2x is a square of that sum or difference, less or plus 1. Rewrite double angles through single ones first so the right factor shows.

Definition

  • An odd power of cos⁡x\cos x: put t=sin⁡xt=\sin x. An odd power of sin⁡x\sin x: put t=cos⁡xt=\cos x.
  • Top cos⁡x−sin⁡x\cos x-\sin x: put t=sin⁡x+cos⁡xt=\sin x+\cos x, and sin⁡2x=t2−1\sin2x=t^2-1.
  • Top cos⁡x+sin⁡x\cos x+\sin x: put t=sin⁡x−cos⁡xt=\sin x-\cos x, and sin⁡2x=1−t2\sin2x=1-t^2.
  • Double angles first: 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x, sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x.

Sum substitution

(sin⁡x±cos⁡x)2=1±sin⁡2x(\sin x\pm\cos x)^2=1\pm\sin2x

Worked example

Find ∫sin⁡x+cos⁡x3+sin⁡2x dx\int\frac{\sin x+\cos x}{3+\sin2x}\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q70Moderate

Example 2 · Indefinite Integration · Trigonometric Integrals

For x∈(−π2,π2)x \in\left( -\frac{\pi}{2},\frac{\pi}{2} \right), if y(x)=∫csc⁡x+sin⁡xcsc⁡xsec⁡x+tan⁡xsin⁡2xdxy(x) = \int\frac{\csc x + \sin x}{\csc x\sec x + \tan x\sin^{2}x}dx and lim⁡x→(π2)−y(x)=0\lim_{x \rightarrow\left( \frac{\pi}{2} \right)^{-}} y(x) = 0 then y(π4)y\left( \frac{\pi}{4} \right) is equal to

Pick t by the top

For cos⁡x−sin⁡x\cos x-\sin x on top, put t=sin⁡x+cos⁡xt=\sin x+\cos x; for cos⁡x+sin⁡x\cos x+\sin x, put t=sin⁡x−cos⁡xt=\sin x-\cos x. The other choice leaves no dtdt.
Drill 4 more on substituting sin x, cos x or sin x ± cos x

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 (2)

  • Everything in tan x or cot x

    Tan substitution

    ∫dxa2sin⁡2x+b2cos⁡2x=1abtan⁡−1(atan⁡xb)+C\int\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac1{ab}\tan^{-1}\left(\frac{a\tan x}{b}\right)+C
  • Substituting sin x, cos x or sin x ± cos x

    Sum substitution

    (sin⁡x±cos⁡x)2=1±sin⁡2x(\sin x\pm\cos x)^2=1\pm\sin2x

Watch out for (2)

Test yourself on Indefinite Integration

15 past JEE Mains questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Indefinite Integration chapter testDrill all 9 questions