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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 by a standard result or one identity, before any heavy substitution. Four reflexes: the standard tan/cot/sec/cosec integrals (recall, not re-derive), power-reduction (turning tan⁴x or sin²x into integrable pieces), identity-simplification (collapsing tan x + cot x, or sin(5x/2)/sin(x/2)), and reducing an inverse-trig argument to a linear function of x. Master these and the harder rational-in-sin/cos integrals in Trigonometric Integrals II become approachable.

Concept 1 of 4

The Standard tan, cot, sec, cosec Integrals

Intuition

Four trig integrals appear so often they are worth knowing as results, not re-deriving each time. ∫tan⁡\int\tan and ∫cot⁡\int\cot are just f′/ff'/f logs; ∫sec⁡\int\sec and ∫csc⁡\int\csc come from a one-time multiply-by-the-conjugate trick. Memorise the four answers.

Definition

The four results, each holding on the domain where the function is defined:

  • ∫tan⁡x dx=log⁡∣sec⁡x∣+C\int \tan x\,dx = \log|\sec x| + C (=−log⁡∣cos⁡x∣= -\log|\cos x|) — top is −(cos⁡x)′-(\cos x)', an f′/ff'/f log
  • ∫cot⁡x dx=log⁡∣sin⁡x∣+C\int \cot x\,dx = \log|\sin x| + C — top is (sin⁡x)′(\sin x)', an f′/ff'/f log
  • ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C\int \sec x\,dx = \log|\sec x + \tan x| + C — from multiplying by sec⁡x+tan⁡xsec⁡x+tan⁡x\dfrac{\sec x + \tan x}{\sec x + \tan x}
  • ∫csc⁡x dx=log⁡∣csc⁡x−cot⁡x∣+C\int \csc x\,dx = \log|\csc x - \cot x| + C — from multiplying by csc⁡x−cot⁡xcsc⁡x−cot⁡x\dfrac{\csc x - \cot x}{\csc x - \cot x}

The two that need the conjugate trick

∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C,∫csc⁡x dx=log⁡∣csc⁡x−cot⁡x∣+C\int \sec x\,dx = \log|\sec x + \tan x| + C,\qquad \int \csc x\,dx = \log|\csc x - \cot x| + C

Worked example

Derive ∫sec⁡x dx\displaystyle\int \sec x\,dx.
  1. Multiply top and bottom by sec⁡x+tan⁡x\sec x + \tan x: ∫sec⁡x(sec⁡x+tan⁡x)sec⁡x+tan⁡x dx=∫sec⁡2x+sec⁡xtan⁡xsec⁡x+tan⁡x dx\displaystyle\int \dfrac{\sec x(\sec x + \tan x)}{\sec x + \tan x}\,dx = \int \dfrac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}\,dx.
  2. The numerator is exactly ddx(sec⁡x+tan⁡x)=sec⁡xtan⁡x+sec⁡2x\dfrac{d}{dx}(\sec x + \tan x) = \sec x\tan x + \sec^2 x — so the integrand is f′/ff'/f.
  3. Therefore ∫f′f dx=log⁡∣f∣=log⁡∣sec⁡x+tan⁡x∣+C\int \dfrac{f'}{f}\,dx = \log|f| = \log|\sec x + \tan x| + C.
Answer:∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C\int \sec x\,dx = \log|\sec x + \tan x| + C
Practice this concept4 quick reps

∫sec and ∫cosec are NOT plain logs of sec/cosec

∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣\int \sec x\,dx = \log|\sec x + \tan x|, not log⁡∣sec⁡x∣\log|\sec x|. The +tan⁡x+\tan x (and −cot⁡x-\cot x for cosec) is exactly what the conjugate trick produces — options drop it to bait you.

Concept 2 of 4

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

From the bank · past-year question

Example 2Indefinite 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 3 of 4

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 (after which the integral is a standard form):

  • 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}
  • ratios like sin⁡(5x/2)sin⁡(x/2)\dfrac{\sin(5x/2)}{\sin(x/2)} expand into a sum of cosines

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

From the bank · past-year question

Example 3Indefinite 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 3 more on identity simplification before integrating

Concept 4 of 4

Simplify the Inverse-Trig Argument First

Intuition

When the integrand is an inverse-trig function OF a trig expression, don't integrate it as written. Use half-angle or double-angle identities to turn the ARGUMENT into tan⁡(linear)\tan(\text{linear}) or sin⁡(linear)\sin(\text{linear}); the inverse and the function then cancel (tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta)=\theta), leaving a linear function of xx you integrate in one line.

Definition

The collapse relies on sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta)=\theta, cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta)=\theta, tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta)=\theta on the principal range. Standard argument reductions:

  • sin⁡2x1+cos⁡2x=tan⁡x\dfrac{\sin 2x}{1+\cos 2x} = \tan x
  • 1−sin⁡x1+sin⁡x=∣tan⁡ ⁣(π4−x2)∣\sqrt{\dfrac{1-\sin x}{1+\sin x}} = \left|\tan\!\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right|
  • 1+cos⁡xsin⁡x=cot⁡x2\dfrac{1+\cos x}{\sin x} = \cot\dfrac{x}{2}, and cos⁡3x=sin⁡ ⁣(π2−3x)\cos 3x = \sin\!\left(\dfrac{\pi}{2}-3x\right)

After the reduction the integrand is linear in xx, so the integral is a simple polynomial.

Cancel, then integrate

tan⁡−1(tan⁡θ)=θ,sin⁡−1(sin⁡θ)=θ(θ in principal range)\tan^{-1}(\tan\theta) = \theta,\quad \sin^{-1}(\sin\theta)=\theta \quad (\theta\text{ in principal range})

Worked example

Evaluate ∫tan⁡−1 ⁣(sin⁡2x1+cos⁡2x)dx\displaystyle\int \tan^{-1}\!\left(\dfrac{\sin 2x}{1+\cos 2x}\right)dx.
  1. Reduce the argument: sin⁡2x1+cos⁡2x=2sin⁡xcos⁡x2cos⁡2x=tan⁡x\dfrac{\sin 2x}{1+\cos 2x} = \dfrac{2\sin x\cos x}{2\cos^2 x} = \tan x.
  2. So the integrand is tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x.
  3. Integrate: ∫x dx=x22+C\int x\,dx = \dfrac{x^2}{2} + C.
Answer:x22+C\dfrac{x^2}{2} + C
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Indefinite 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]

Reduce the argument BEFORE integrating

Never reach for by-parts on ∫tan⁡−1(⋯ ) dx\int\tan^{-1}(\cdots)\,dx until you've tried to collapse the inside. If the argument is a half/double-angle form, the inverse cancels and the integral is a one-line polynomial — by-parts is a needless detour.

Summary — formulas & gotchas at a glance

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Formulas (4)

  • The Standard tan, cot, sec, cosec Integrals

    The two that need the conjugate trick

    ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C,∫csc⁡x dx=log⁡∣csc⁡x−cot⁡x∣+C\int \sec x\,dx = \log|\sec x + \tan x| + C,\qquad \int \csc x\,dx = \log|\csc x - \cot x| + C
  • Power Reduction with Pythagorean Identities

    Key reduction identity

    tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1
  • Identity Simplification before Integrating

    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
  • Simplify the Inverse-Trig Argument First

    Cancel, then integrate

    tan⁡−1(tan⁡θ)=θ,sin⁡−1(sin⁡θ)=θ(θ in principal range)\tan^{-1}(\tan\theta) = \theta,\quad \sin^{-1}(\sin\theta)=\theta \quad (\theta\text{ in principal range})

Watch out for (4)

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