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MHT-CET Maths · Indefinite Integration

Trigonometric Integrals I — Powers and Identities

Rewrite powers and sums of trig functions using identities until what remains is a standard integral.

All Indefinite Integration notesMHT-CET Maths strategy

Why this matters

7 PYQs of the chapter's ~66 trig integrals live here — the ones solved purely by an identity, before any substitution. Power-reduction (turning tan⁴x or sin²x into integrable pieces) and identity-simplification (collapsing tan x + cot x, or sin(5x/2)/sin(x/2)) are the two reflexes. Master these and the harder rational-in-sin/cos integrals in the next subtopic become approachable.

Concept 1 of 2

Power Reduction with Pythagorean Identities

Intuition

A high power of tan or sec is peeled down using tan⁡2=sec⁡2−1\tan^2 = \sec^2 - 1 (or sin⁡2=1−cos⁡2\sin^2 = 1 - \cos^2) until every piece is either a standard integral or a [f]nf′[f]^n f' shape.

Definition

Use the Pythagorean identities sec⁡2x=1+tan⁡2x\sec^2 x = 1 + \tan^2 x and csc⁡2x=1+cot⁡2x\csc^2 x = 1 + \cot^2 x to split high powers. For tan⁡nx\tan^n x, repeatedly write tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1: the sec⁡2x\sec^2 x factor pairs with a tan⁡\tan-power as [tan⁡x]ksec⁡2x[\tan x]^k \sec^2 x (integrates by the power rule), and the leftover lowers the degree.

Key reduction identity

tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1

Worked example

Evaluate ∫cot⁡4x dx\displaystyle\int \cot^4 x\,dx.
  1. Write cot⁡4x=cot⁡2x(csc⁡2x−1)=cot⁡2xcsc⁡2x−cot⁡2x\cot^4 x = \cot^2 x(\csc^2 x - 1) = \cot^2 x\csc^2 x - \cot^2 x.
  2. Replace the leftover: −cot⁡2x=−(csc⁡2x−1)=−csc⁡2x+1-\cot^2 x = -(\csc^2 x - 1) = -\csc^2 x + 1.
  3. Integrate: ∫cot⁡2xcsc⁡2x dx=−cot⁡3x3\int \cot^2 x\csc^2 x\,dx = -\dfrac{\cot^3 x}{3} (u=cot⁡xu = \cot x); ∫−csc⁡2x dx=cot⁡x\int -\csc^2 x\,dx = \cot x; ∫1 dx=x\int 1\,dx = x.
  4. So ∫cot⁡4x dx=−cot⁡3x3+cot⁡x+x+C\int \cot^4 x\,dx = -\dfrac{\cot^3 x}{3} + \cot x + x + C.
Answer:−cot⁡3x3+cot⁡x+x+C-\dfrac{\cot^3 x}{3} + \cot x + x + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate ∫tan⁡2x dx\displaystyle\int \tan^2 x\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    ∫sec⁡2x dx\int \sec^2 x\,dx
  2. 2.
    ∫(sec⁡2x−1) dx\int (\sec^2 x - 1)\,dx
  3. 3.
    ∫tan⁡2xsec⁡2x dx\int \tan^2 x\sec^2 x\,dx
  4. 4.
    ∫cot⁡2x dx\int \cot^2 x\,dx

From the bank · past-year question

Example 1Indefinite IntegrationMODERATE
If ∫tan⁡4x dx=atan⁡3x+btan⁡x+cx+k\int \tan^4 x\,dx = a\tan^3 x + b\tan x + cx + k (where k is the constant of integration) then the value of a−b+c=a - b + c =

[Shift || · 2025]

Keep one sec²x to pair with the tan-power

The whole method works because sec⁡2x dx=d(tan⁡x)\sec^2 x\,dx = d(\tan x). Each reduction must leave a sec⁡2x\sec^2 x attached to a power of tan⁡x\tan x so the power rule applies — otherwise you stall.
Drill 1 more on power reduction with pythagorean identities

Concept 2 of 2

Identity Simplification before Integrating

Intuition

Many trig integrands look exotic but collapse to something standard after ONE identity — a half-angle, a sum-to-product, or just writing everything over sin⁡\sin and cos⁡\cos. Always try to simplify before substituting.

Definition

Common collapses: tan⁡x+cot⁡x=1sin⁡xcos⁡x=2csc⁡2x\tan x + \cot x = \dfrac{1}{\sin x\cos x} = 2\csc 2x; 11+cos⁡x=12sec⁡2x2\dfrac{1}{1+\cos x} = \dfrac{1}{2}\sec^2\dfrac{x}{2}; and ratios like sin⁡(5x/2)sin⁡(x/2)\dfrac{\sin(5x/2)}{\sin(x/2)} expand into a sum of cosines. After the identity, the integral is a standard form.

A workhorse collapse

tan⁡x+cot⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x=2sin⁡2x=2csc⁡2x\tan x + \cot x = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{2}{\sin 2x} = 2\csc 2x

Worked example

Evaluate ∫2cos⁡2x dx\displaystyle\int 2\cos^2 x\,dx.
  1. Use the double-angle identity 2cos⁡2x=1+cos⁡2x2\cos^2 x = 1 + \cos 2x to collapse the power.
  2. The integrand is now a standard sum: ∫(1+cos⁡2x) dx\int (1 + \cos 2x)\,dx.
  3. Integrate: x+sin⁡2x2+Cx + \dfrac{\sin 2x}{2} + C.
Answer:x+sin⁡2x2+Cx + \dfrac{\sin 2x}{2} + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate ∫(1−cos⁡x)csc⁡2x dx\displaystyle\int (1 - \cos x)\csc^2 x\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Simplify tan⁡x+cot⁡x\tan x + \cot x.
  2. 2.
    ∫dx1+cos⁡x\int \dfrac{dx}{1+\cos x}
  3. 3.
    ∫2csc⁡2x dx\int 2\csc 2x\,dx
  4. 4.
    ∫dx1−cos⁡x\int \dfrac{dx}{1-\cos x}

From the bank · past-year question

Example 2Indefinite IntegrationMODERATE
∫(tan⁡x+cot⁡x) dx=\int(\tan x + \cot x)\,dx =

[Q131 · 9th May Shift 2 · 2024]

Try an identity before a substitution

Reaching for Weierstrass on ∫(tan⁡x+cot⁡x) dx\int(\tan x + \cot x)\,dx is a long detour — one identity makes it 2csc⁡2x2\csc 2x instantly. Simplify first; substitute only if no identity collapses it.
Drill 4 more on identity simplification before integrating

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)

Watch out for (2)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Indefinite IntegrationMODERATE
If f(x)=tan⁡xf(x) = \tan x and g(x)=sin⁡x⋅cos⁡xg(x) = \sin x \cdot \cos x then ∫f(x) g(x) dx\int f(x)\,g(x)\,dx is equal to (where C is a constant of integration)

[Q108 · Shift 1 · 2022]

Example 2Indefinite IntegrationMODERATE
∫x+sin⁡x1+cos⁡x dx=\int \frac{x + \sin x}{1 + \cos x}\,dx =

[Shift || · 2025]

Example 3Indefinite IntegrationMODERATE
∫tan⁡−1 ⁣1−sin⁡x1+sin⁡x dx=\int\tan^{-1}\!\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx=

[Q142 · 4th May Shift 2 · 2023]

Example 4Indefinite IntegrationMODERATE
The value of ∫(1−cos⁡x)csc⁡2x dx\int(1-\cos x)\csc^2 x\,dx is

[Q109 · 11th May Shift 2 · 2024]

Example 5Indefinite IntegrationHARD
∫sin⁡5x2sin⁡x2 dx=\int \frac{\sin\frac{5x}{2}}{\sin\frac{x}{2}}\,dx = (where C is a constant of integration.)

[Q147 · Shift 1 · 2022]

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