MHT-CET Maths · Indefinite Integration

Trigonometric Integrals II — Rational Forms and Substitutions

The hard trig core — fractions in sine and cosine, handled by the half-angle (Weierstrass) substitution, the divide-by-cosine-squared move, the fractional-power tangent trick, and numerator-matching.

All Indefinite Integration notes MHT-CET Maths strategy

Why this matters

26 PYQs and the chapter's HARDEST pocket — 25 of the 26 are HARD. These are the integrals that decide a top score. Four named techniques cover almost all of them: Weierstrass t = tan(x/2) for 1/(a+b sin x); divide-by-cos² for 1/(a+b sin²x); the fractional-power tan trick for cos/sin power products; and writing a numerator as 'denominator + its derivative'. Learn to RECOGNISE which of the four a question wants — that recognition is the whole skill.

Concept 1 of 4

The Half-Angle (Weierstrass) Substitution

Intuition

Any rational function of

\sin x

and

\cos x

becomes a rational function of one variable

t = \tan(x/2)

. It is the universal hammer for

\dfrac{1}{a + b\sin x}

and

\dfrac{1}{a + b\cos x}

Definition

With $t = \tan\dfrac{x}{2}$ : $\sin x = \dfrac{2t}{1+t^2}$ , $\cos x = \dfrac{1-t^2}{1+t^2}$ , and $dx = \dfrac{2\,dt}{1+t^2}$ . Substituting turns the integral into a rational function of $t$ , finished by completing the square and an arctan.

Weierstrass substitution

t = \tan\dfrac{x}{2}:\quad \sin x = \dfrac{2t}{1+t^2},\ \cos x = \dfrac{1-t^2}{1+t^2},\ dx = \dfrac{2\,dt}{1+t^2}

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{5 + 3\cos x}

Substitute $t = \tan(x/2)$ : $\cos x = \dfrac{1-t^2}{1+t^2}$ , $dx = \dfrac{2\,dt}{1+t^2}$ .
Denominator: $5 + 3\cdot\dfrac{1-t^2}{1+t^2} = \dfrac{5(1+t^2) + 3(1-t^2)}{1+t^2} = \dfrac{8 + 2t^2}{1+t^2}$ .
Integral becomes $\int \dfrac{2\,dt}{8 + 2t^2} = \int \dfrac{dt}{4 + t^2} = \dfrac{1}{2}\tan^{-1}\dfrac{t}{2}$ (the cosine form needs no completing-the-square — there is no linear $t$ term).
Back-substitute $t = \tan(x/2)$ .

Answer:

\dfrac{1}{2}\tan^{-1}\!\left(\dfrac{\tan(x/2)}{2}\right) + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Set up

\displaystyle\int \dfrac{dx}{1 + \sin x}

with the half-angle substitution.

Practice — Level 1 (4 reps)

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

1.
Under $t=\tan(x/2)$ , what is $\sin x$ ?
2.
Under $t=\tan(x/2)$ , what is $dx$ ?
3.
Under $t=\tan(x/2)$ , what is $\cos x$ ?
4.
$\int \dfrac{dx}{1+\sin x}$

From the bank · past-year question

Example 1Indefinite IntegrationHARD

The value of

\int \frac{dx}{5+4\sin x}

is equal to

[Q122 · 15th May Shift 1 · 2023]

Weierstrass is for a + b·sin/cos, not a + b·sin²

If the denominator has

\sin^2 x

\cos^2 x

(an even power), the half-angle substitution gives a messy quartic. Use divide-by-

\cos^2 x

instead — the next concept.

Drill 6 more on the half-angle (weierstrass) substitution

Concept 2 of 4

Divide by cos²x for a + b·sin²x Forms

Intuition

When the denominator is built from

\sin^2 x

and

\cos^2 x

, divide top and bottom by

\cos^2 x

. Everything turns into

\tan x

and

\sec^2 x

, and

t = \tan x

finishes it as an arctan.

Definition

For $\displaystyle\int \dfrac{dx}{a + b\sin^2 x}$ (or with $\cos^2 x$ ): divide numerator and denominator by $\cos^2 x$ , using $\dfrac{1}{\cos^2 x} = \sec^2 x$ and $\dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x$ . Then $t = \tan x,\ dt = \sec^2 x\,dx$ gives $\int \dfrac{dt}{A + Bt^2}$ , a standard arctan.

After dividing by cos²x

\int \dfrac{\sec^2 x\,dx}{A + B\tan^2 x} \xrightarrow{\,t=\tan x\,} \int \dfrac{dt}{A + Bt^2}

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{4\cos^2 x + 9\sin^2 x}

Divide top and bottom by $\cos^2 x$ : $\dfrac{\sec^2 x}{4 + 9\tan^2 x}$ .
Let $t = \tan x$ , $dt = \sec^2 x\,dx$ : the integral becomes $\int \dfrac{dt}{4 + 9t^2}$ .
Standard arctan-quadratic form: $\int \dfrac{dt}{4 + 9t^2} = \dfrac{1}{6}\tan^{-1}\dfrac{3t}{2}$ .
Back-substitute $t = \tan x$ .

Answer:

\dfrac{1}{6}\tan^{-1}\!\left(\dfrac{3\tan x}{2}\right) + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{dx}{1 + \tan^2 x}

using this idea.

Practice — Level 1 (4 reps)

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

1.
Divide $\dfrac{1}{a+b\sin^2 x}$ by $\cos^2 x$ : numerator becomes?
2.
$\int \dfrac{\sec^2 x}{1 + \tan^2 x}\,dx$
3.
After $t=\tan x$ , $\int \dfrac{dt}{1+4t^2}$ ?
4.
$\int \dfrac{dx}{\cos^2 x(1+\tan^2 x)}$ — simplify the integrand.

From the bank · past-year question

Example 2Indefinite IntegrationHARD

\int \frac{dx}{1+3\sin^{2}x} = \frac{1}{2}\tan^{-1}(f(x))+c

, where c is a constant of integration, then

f(x)

is equal to

[Q105 · 15th May Shift 1 · 2023]

Even powers → divide by cos²; odd → Weierstrass

Denominator has

\sin^2/\cos^2

(even): divide by

\cos^2 x

, use

t=\tan x

. Denominator has plain

\sin x/\cos x

(odd): use Weierstrass

t=\tan(x/2)

. Picking the wrong one makes the algebra explode.

Drill 2 more on divide by cos²x for a + b·sin²x forms

Concept 3 of 4

The Fractional-Power tan Trick

Intuition

For a product of powers of

\sin x

and

\cos x

whose exponents add up to an even negative integer, factoring out

\sec^2 x

and substituting

t = \tan x

turns the whole thing into a plain power of

t

— even when the exponents are fractions.

Definition

If the integrand is $\sin^m x\,\cos^n x$ with $m + n$ an even negative integer, write it as a power of $\tan x$ times $\sec^2 x$ : $\sin^m x\cos^n x = \tan^m x\cdot\cos^{m+n}x = (\tan x)^m(\sec^2 x)^{-(m+n)/2}$ . Then $t = \tan x$ reduces it to $\int t^m(1+t^2)^{-(m+n)/2 - 1}$ -type powers — often a single $\int t^p\,dt$ .

The reduction (m + n even)

\int \sin^m x\,\cos^n x\,dx \xrightarrow{\,t=\tan x\,} \int t^{m}\,(1+t^2)^{\frac{m+n}{2}-1}\,dt

Worked example

Evaluate

\displaystyle\int \sin^{-5/3}x\,\cos^{-1/3}x\,dx

Exponents: $m = -\tfrac{5}{3}$ (sin), $n = -\tfrac{1}{3}$ (cos); sum $m+n = -2$ , an even negative integer — the trick applies.
Rewrite: $\sin^{-5/3}x\,\cos^{-1/3}x = \tan^{-5/3}x\cdot\cos^{-2}x = \tan^{-5/3}x\,\sec^2 x$ .
Let $t = \tan x,\ dt = \sec^2 x\,dx$ : $\int t^{-5/3}\,dt = \dfrac{t^{-2/3}}{-2/3} = -\dfrac{3}{2}\,t^{-2/3}$ .
Back-substitute $t = \tan x$ .

Answer:

-\dfrac{3}{2}(\tan x)^{-2/3} + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \sec^{2/3}x\,\csc^{4/3}x\,dx

by the same route.

Practice — Level 1 (4 reps)

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

1.
For $\sin^m x\cos^n x$ , the trick needs $m+n$ to be?
2.
Rewrite $\sin^{-1/2}x\cos^{-3/2}x$ (sum $=-2$ ).
3.
$\int \tan^{-1/2}x\,\sec^2 x\,dx$
4.
$\int \tan^{3/2}x\,\sec^2 x\,dx$

From the bank · past-year question

Example 3Indefinite IntegrationHARD

\int \cos^{-3/7}x \cdot \sin^{-11/7}x\,dx =

[Q128 · Shift 1 · 2023]

Check m + n is an even integer first

The trick only collapses cleanly when

m + n

is an even integer (so

\cos^{m+n}x

becomes an integer power of

\sec^2 x

). If it is odd, this route leaves a stray

\sec x

\cos x

and you need a different method.

Drill 9 more on the fractional-power tan trick

Concept 4 of 4

Numerator as Denominator + its Derivative

Intuition

For

\dfrac{a\sin x + b\cos x}{c\sin x + d\cos x}

-type fractions, write the numerator as a combination of the denominator and the denominator's derivative. The first piece integrates to

x

-times-a-constant, the second to a clean log.

Definition

Express $\text{numerator} = A\,(\text{denominator}) + B\,\dfrac{d}{dx}(\text{denominator})$ . Then $\displaystyle\int \dfrac{\text{num}}{\text{den}}\,dx = A\!\int 1\,dx + B\!\int \dfrac{(\text{den})'}{\text{den}}\,dx = Ax + B\log|\text{den}| + C$ . Solve for $A, B$ by matching the $\sin x$ and $\cos x$ coefficients.

Decomposition of the numerator

\text{num} = A\cdot\text{den} + B\cdot(\text{den})' \;\Rightarrow\; \int\dfrac{\text{num}}{\text{den}}\,dx = Ax + B\log|\text{den}| + C

Worked example

Evaluate

\displaystyle\int \dfrac{2\sin x + 3\cos x}{\sin x + \cos x}\,dx

Denominator $D = \sin x + \cos x$ , so $D' = \cos x - \sin x$ . Write $2\sin x + 3\cos x = A\,D + B\,D'$ .
Match coefficients — $\sin x$ : $A - B = 2$ ; $\cos x$ : $A + B = 3$ . Solve: $A = \tfrac52,\ B = \tfrac12$ .
Integrate: $\int\!\left(A + B\dfrac{D'}{D}\right)dx = \dfrac{5}{2}x + \dfrac{1}{2}\log|\sin x + \cos x| + C$ .

Answer:

\dfrac{5}{2}x + \dfrac{1}{2}\log|\sin x + \cos x| + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{dx}{\sin x + \cos x}

Practice — Level 1 (4 reps)

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

1.
Derivative of $\sin x - 2\cos x$ ?
2.
Write $\sin x + \cos x$ as a single sine.
3.
$\int \dfrac{(\sin x - 2\cos x)'}{\sin x - 2\cos x}\,dx$
4.
If num $= 1\cdot D + 2\cdot D'$ , the integral is?

From the bank · past-year question

Example 4Indefinite IntegrationHARD

\displaystyle\int\frac{5\tan x}{\tan x - 2}\,dx = x + a\log|\sin x - 2\cos x| + c

, then the value of

a

[Q111 · 10th May Shift 2 · 2024]

Convert tan-fractions to sin/cos first

A ratio in

\tan x

is easiest after multiplying through by

\cos x

to get a sin/cos ratio — then the 'denominator + its derivative' decomposition is clean.

Drill 5 more on numerator as denominator + its derivative

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 Half-Angle (Weierstrass) Substitution
Weierstrass substitution
$t = \tan\dfrac{x}{2}:\quad \sin x = \dfrac{2t}{1+t^2},\ \cos x = \dfrac{1-t^2}{1+t^2},\ dx = \dfrac{2\,dt}{1+t^2}$
Divide by cos²x for a + b·sin²x Forms
After dividing by cos²x
$\int \dfrac{\sec^2 x\,dx}{A + B\tan^2 x} \xrightarrow{\,t=\tan x\,} \int \dfrac{dt}{A + Bt^2}$
The Fractional-Power tan Trick
The reduction (m + n even)
$\int \sin^m x\,\cos^n x\,dx \xrightarrow{\,t=\tan x\,} \int t^{m}\,(1+t^2)^{\frac{m+n}{2}-1}\,dt$
Numerator as Denominator + its Derivative
Decomposition of the numerator
$\text{num} = A\cdot\text{den} + B\cdot(\text{den})' \;\Rightarrow\; \int\dfrac{\text{num}}{\text{den}}\,dx = Ax + B\log|\text{den}| + C$

Watch out for (4)

Weierstrass is for a + b·sin/cos, not a + b·sin²
→ The Half-Angle (Weierstrass) Substitution
Even powers → divide by cos²; odd → Weierstrass
→ Divide by cos²x for a + b·sin²x Forms
Check m + n is an even integer first
→ The Fractional-Power tan Trick
Convert tan-fractions to sin/cos first
→ Numerator as Denominator + its Derivative

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

The value of

\int\frac{\cos^3 x}{\sin^2 x + \sin x}\,dx

[Q145 · 11th May Shift 1 · 2023]

Example 2Indefinite IntegrationHARD

\int\dfrac{\sin x}{3 + 4\cos^2 x}\,dx = A\tan^{-1}(B\cos x) + c

, (where

c

is a constant of integration), then the value of

A + B

[Q124 · 9th May Shift 2 · 2024]

Example 3Indefinite IntegrationHARD

\int\frac{dx}{\cos^3 x\sqrt{2\sin2x}}=(\tan x)^A+\frac{C}{(\tan x)^B}+K

, where K is a constant of integration, then the value of

5(A+B+C)

is equal to

[Q101 · 3rd May 2nd Shift · 2023]

Example 4Indefinite IntegrationHARD

I=\int\frac{\sin x+\sin^3 x}{\cos 2x}\,dx=P\cos x+Q\log\left|\frac{2\cos x-1}{2\cos x+1}\right|+C

, then the values of

P

and

Q

are respectively

[Q106 · 9th May Shift 1 · 2024]

Example 5Indefinite IntegrationHARD

\int \text{cosec}(x-a) \cdot \text{cosec}\,x\, dx =

[Q106 · 15th May Shift 2 · 2023]

Drill every past-year question on this subtopic

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