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

46 PYQs and the chapter's HARDEST pocket — 35 of the 46 are HARD. These are the integrals that decide a top score, and they lean on the standard arctan/log forms you met in Rational Functions & Partial Fractions. Six 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) (and tan x = t for the 2x version); the product-of-sines split for 1/(sin(x−a)sin(x−b)); the trig-to-partial-fraction bridge when a substitution makes it rational; the fractional-power tan trick for cos/sin power products; and writing a numerator as 'denominator + its derivative'. Learn to RECOGNISE which one a question wants — that recognition is the whole skill.

Concept 1 of 6: The Half-Angle (Weierstrass) Substitution

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}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q122Hard

Example 1 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

The value of

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

is equal to

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 1 more on the half-angle (weierstrass) substitution

Concept 2 of 6: Divide by cos²x for a + b·sin²x Forms

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. Double-angle version: for $\displaystyle\int \dfrac{dx}{a\sin 2x + b\cos 2x + c}$ , substitute $t = \tan x$ directly — then $\sin 2x = \dfrac{2t}{1+t^2}$ , $\cos 2x = \dfrac{1-t^2}{1+t^2}$ , $dx = \dfrac{dt}{1+t^2}$ , and the integral reduces to $\int \dfrac{dt}{At^2+Bt+C}$ . This is the $2x$ analogue of Weierstrass (which uses $t=\tan\tfrac{x}{2}$ for the plain-angle case).

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}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q105Hard

Example 2 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

\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

Three sin/cos denominators, three substitutions

Denominator has

\sin^2/\cos^2

(even, plain angle): divide by

\cos^2 x

, use

t=\tan x

. Denominator has

\sin 2x/\cos 2x

(double angle): use

t=\tan x

with the

2x

formulae. Denominator has plain

\sin x/\cos x

(odd, single power): 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 6: Product of Two Shifted Sines (or Cosines)

For

\dfrac{1}{\sin(x-a)\sin(x-b)}

, don't substitute — split it. Writing the constant

\sin(a-b)

\sin

of the angle DIFFERENCE turns the product into a difference of cotangents, each a clean log. One identity, one line.

Definition

Use $\sin\big((x-a)-(x-b)\big) = \sin(b-a)$ , a constant. Expanding and dividing by $\sin(x-a)\sin(x-b)$ gives $\dfrac{1}{\sin(x-a)\sin(x-b)} = \dfrac{1}{\sin(b-a)}\big[\cot(x-b) - \cot(x-a)\big]$ , which integrates to a difference of $\log|\sin|$ terms. The same idea handles $\dfrac{1}{\cos(x-a)\cos(x-b)}$ (difference of $\tan$ s) and $\dfrac{1}{\sin(x-a)\cos(x-b)}$ .

Shifted-angle split

\dfrac{1}{\sin(x-a)\sin(x-b)} = \dfrac{1}{\sin(b-a)}\big[\cot(x-b) - \cot(x-a)\big]

$\sin(b-a)$ constant divisor — the sine of the difference of the two shifts

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{\sin x\,\sin(x+a)}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q111Hard

Example 3 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

I = \int \frac{dx}{\sin(x-a)\sin(x-b)}

, then

I

is given by

Split it — don't reach for Weierstrass

Weierstrass on a product of two shifted sines produces a quartic in

t

and stalls. The shifted-angle split is one identity and one line. Keep the constant

\sin(b-a)

out front — it is NOT a function of

x

Drill 2 more on product of two shifted sines (or cosines)

Concept 4 of 6: Trig to Partial Fractions (substitute, then decompose)

A trig integrand that becomes a RATIONAL function after one substitution hands straight off to the partial-fraction machinery from the previous movement. When there's a spare

\cos x\,dx

(

=d(\sin x)

) or

\sin x\,dx

(

=-d(\cos x)

), substitute and decompose.

Definition

Spot the bridge: if substituting $t=\sin x$ (needs a spare $\cos x\,dx$ ) or $t=\cos x$ (needs a spare $\sin x\,dx$ ) leaves a rational function of $t$ , do it — then split by partial fractions and integrate each piece as a log/arctan. An odd power of the 'spare' function is the signal: e.g. $\cos^3 x = \cos x(1-\sin^2 x)$ frees one $\cos x\,dx$ for $t=\sin x$ . Rewrite any $\cos^2 x$ as $1-\sin^2 x$ (or $\sin^2 x$ as $1-\cos^2 x$ ) so the rest is rational in $t$ .

The bridge

\int R(\sin x)\,\cos x\,dx \;\xrightarrow{\,t=\sin x\,}\; \int R(t)\,dt \;\;(\text{now rational — decompose})

Worked example

Evaluate

\displaystyle\int \dfrac{\cos x}{(1+\sin x)(3+\sin x)}\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q145Moderate

Example 4 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

The value of

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

No spare cos/sin → no bridge

The substitution only works if exactly one

\cos x\,dx

(or

\sin x\,dx

) is available to become

dt

and everything else turns rational in

t

. If both

\sin x

and

\cos x

appear in odd/mixed ways with nothing spare, fall back to Weierstrass.

Drill 1 more on trig to partial fractions (substitute, then decompose)

Concept 5 of 6: The Fractional-Power tan Trick

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q128Hard

Example 5 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

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

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 7 more on the fractional-power tan trick

Concept 6 of 6: Numerator as Denominator + its Derivative

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q111Hard

Example 6 · Indefinite Integration · Trigonometric Integrals - Rational and Substitution Forms

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

, then the value of

a

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 (6)

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
$\int \dfrac{\sec^2 x\,dx}{A + B\tan^2 x} \xrightarrow{\,t=\tan x\,} \int \dfrac{dt}{A + Bt^2}$
Product of Two Shifted Sines (or Cosines)
Shifted-angle split
$\dfrac{1}{\sin(x-a)\sin(x-b)} = \dfrac{1}{\sin(b-a)}\big[\cot(x-b) - \cot(x-a)\big]$
Trig to Partial Fractions (substitute, then decompose)
The bridge
$\int R(\sin x)\,\cos x\,dx \;\xrightarrow{\,t=\sin x\,}\; \int R(t)\,dt \;\;(\text{now rational — decompose})$
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 (6)

Weierstrass is for a + b·sin/cos, not a + b·sin²
→ The Half-Angle (Weierstrass) Substitution
Three sin/cos denominators, three substitutions
→ Divide by cos²x for a + b·sin²x Forms
Split it — don't reach for Weierstrass
→ Product of Two Shifted Sines (or Cosines)
No spare cos/sin → no bridge
→ Trig to Partial Fractions (substitute, then decompose)
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

Test yourself on Indefinite Integration

20 past MHT-CET 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 test Drill all 46 questions

Test yourself on Indefinite Integration

Related notes