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, 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

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

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 1 more on the half-angle (weierstrass) substitution

Concept 2 of 6

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. 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}

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

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]

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)

Intuition

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)}

Write $\sin a = \sin\big((x+a)-x\big) = \sin(x+a)\cos x - \cos(x+a)\sin x$ .
Divide by $\sin x\,\sin(x+a)$ : $\dfrac{\sin a}{\sin x\,\sin(x+a)} = \cot x - \cot(x+a)$ .
So the integrand is $\dfrac{1}{\sin a}\big[\cot x - \cot(x+a)\big]$ .
Integrate: $\dfrac{1}{\sin a}\big[\log|\sin x| - \log|\sin(x+a)|\big]$ .

Answer:

\dfrac{1}{\sin a}\log\left|\dfrac{\sin x}{\sin(x+a)}\right| + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Indefinite IntegrationHARD

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

, then

I

is given by

[Q111 · 14th May Shift 2 · 2024]

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)

Intuition

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

Substitute $t = \sin x$ , $dt = \cos x\,dx$ : the integral becomes $\int \dfrac{dt}{(1+t)(3+t)}$ .
Partial fractions: $\dfrac{1}{(1+t)(3+t)} = \dfrac{1}{2}\!\left(\dfrac{1}{1+t} - \dfrac{1}{3+t}\right)$ .
Integrate: $\dfrac{1}{2}\big[\log|1+t| - \log|3+t|\big]$ . Back-substitute $t=\sin x$ .

Answer:

\dfrac{1}{2}\log\left|\dfrac{1+\sin x}{3+\sin x}\right| + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Indefinite IntegrationMODERATE

The value of

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

[Q145 · 11th May Shift 1 · 2023]

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

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

From the bank · past-year question

Example 5Indefinite 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 6 of 6

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

From the bank · past-year question

Example 6Indefinite 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 (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

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