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 notes MHT-CET Maths strategy

Why this matters

19 PYQs of the chapter's 65 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

Four trig integrals appear so often they are worth knowing as results, not re-deriving each time.

\int\tan

and

\int\cot

are just

f'/f

logs;

\int\sec

and

\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:

$\int \tan x\,dx = \log|\sec x| + C$ ( $= -\log|\cos x|$ ) — top is $-(\cos x)'$ , an $f'/f$ log
$\int \cot x\,dx = \log|\sin x| + C$ — top is $(\sin x)'$ , an $f'/f$ log
$\int \sec x\,dx = \log|\sec x + \tan x| + C$ — from multiplying by $\dfrac{\sec x + \tan x}{\sec x + \tan x}$
$\int \csc x\,dx = \log|\csc x - \cot x| + C$ — from multiplying by $\dfrac{\csc x - \cot x}{\csc x - \cot x}$

The two that need the conjugate trick

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

Worked example

Derive

\displaystyle\int \sec x\,dx

Practice this concept4 quick reps

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

\int \sec x\,dx = \log|\sec x + \tan x|

, not

\log|\sec x|

. The

+\tan x

(and

-\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

A high power of tan or sec is peeled down using

\tan^2 = \sec^2 - 1

(or

\sin^2 = 1 - \cos^2

) until every piece is either a standard integral or a

[f]^n f'

shape.

Definition

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

Key reduction identity

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

Worked example

Evaluate

\displaystyle\int \cot^4 x\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q141Moderate

Example 2 · Indefinite Integration · Trigonometric Integrals - Powers and Identities

\int_{}^{}\ \tan^{4}x\text{ }dx= a\tan^{3}x+ btanx+ cx+ k

(where k is the constant of integration) then the value of

a - b + c =

Keep one sec²x to pair with the tan-power

The whole method works because

\sec^2 x\,dx = d(\tan x)

. Each reduction must leave a

\sec^2 x

attached to a power of

\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

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

and

\cos

. Always try to simplify before substituting.

Definition

Common collapses (after which the integral is a standard form):

$\tan x + \cot x = \dfrac{1}{\sin x\cos x} = 2\csc 2x$
$\dfrac{1}{1+\cos x} = \dfrac{1}{2}\sec^2\dfrac{x}{2}$
ratios like $\dfrac{\sin(5x/2)}{\sin(x/2)}$ expand into a sum of cosines

A workhorse collapse

\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

\displaystyle\int 2\cos^2 x\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q131Moderate

Example 3 · Indefinite Integration · Trigonometric Integrals - Powers and Identities

\int(\tan x + \cot x)\,dx =

Try an identity before a substitution

Reaching for Weierstrass on

\int(\tan x + \cot x)\,dx

is a long detour — one identity makes it

2\csc 2x

instantly. Simplify first; substitute only if no identity collapses it.

Drill 2 more on identity simplification before integrating

Concept 4 of 4: Simplify the Inverse-Trig Argument First

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(\text{linear})

\sin(\text{linear})

; the inverse and the function then cancel (

\tan^{-1}(\tan\theta)=\theta

), leaving a linear function of

x

you integrate in one line.

Definition

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

$\dfrac{\sin 2x}{1+\cos 2x} = \tan x$
$\sqrt{\dfrac{1-\sin x}{1+\sin x}} = \left|\tan\!\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right|$
$\dfrac{1+\cos x}{\sin x} = \cot\dfrac{x}{2}$ , and $\cos 3x = \sin\!\left(\dfrac{\pi}{2}-3x\right)$

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

Cancel, then integrate

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

Worked example

Evaluate

\displaystyle\int \tan^{-1}\!\left(\dfrac{\sin 2x}{1+\cos 2x}\right)dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q142Moderate

Example 4 · Indefinite Integration · Trigonometric Integrals - Powers and Identities

\int\tan^{-1}\!\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx=

Reduce the argument BEFORE integrating

Never reach for by-parts on

\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

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

Formulas (4)

The Standard tan, cot, sec, cosec Integrals
The two that need the conjugate trick
$\int \sec x\,dx = \log|\sec x + \tan x| + C,\qquad \int \csc x\,dx = \log|\csc x - \cot x| + C$
Power Reduction with Pythagorean Identities
Key reduction identity
$\tan^2 x = \sec^2 x - 1$
Identity Simplification before Integrating
A workhorse collapse
$\tan x + \cot x = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{2}{\sin 2x} = 2\csc 2x$
Simplify the Inverse-Trig Argument First
Cancel, then integrate
$\tan^{-1}(\tan\theta) = \theta,\quad \sin^{-1}(\sin\theta)=\theta \quad (\theta\text{ in principal range})$

Watch out for (4)

∫sec and ∫cosec are NOT plain logs of sec/cosec
→ The Standard tan, cot, sec, cosec Integrals
Keep one sec²x to pair with the tan-power
→ Power Reduction with Pythagorean Identities
Try an identity before a substitution
→ Identity Simplification before Integrating
Reduce the argument BEFORE integrating
→ Simplify the Inverse-Trig Argument First

Test yourself on Indefinite Integration

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Related notes

Trigonometric Integrals II — rational-in-sin/cos and the half-angle substitution →