If $\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$, where C is the integration constant, then AB is equal to

(D) $8 \operatorname{cosec}(2 \theta)$

(A) $4 \operatorname{cosec}(2 \theta)$

Indefinite Integrals — JEE Main · 2024 · 29 Jan 2024 · Shift 2 Question with Solution

\int \frac{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C

,
where C is the integration constant, then AB is equal to

Answer: (D)

Since

sin (x - θ) = sin x cos θ - cos x sin θ

, factor it two ways:

sin (x - θ) = cos x (tan x cos θ - sin θ) = sin x (cos θ - sin θ cot x)

.

Split the integrand into two pieces,

I = I_{1} + I_{2}

, using

sin^{3} x cos^{3} x sin (x - θ)

under the root.

I_{1} = \int \frac{sin ^{3/2} x}{sin ^{3/2} x cos ^{3/2} x sin ( x - θ )} d x = \int \frac{d x}{cos ^{3/2} x cos x tan x cos θ - sin θ} = \int \frac{sec ^{2} x d x}{cos θ tan x - sin θ}

.

Substituting

t = cos θ tan x - sin θ, d t = cos θ sec^{2} x d x

I_{1} = \frac{1}{cos θ} \int t^{- 1/2} d t = \frac{2 t}{cos θ} = 2 sec θ cos θ tan x - sin θ

.

So

A = 2 sec θ

I_{2} = \int \frac{cos ^{3/2} x}{sin ^{3/2} x cos ^{3/2} x sin ( x - θ )} d x = \int \frac{d x}{sin ^{3/2} x sin x cos θ - sin θ cot x} = \int \frac{csc ^{2} x d x}{cos θ - sin θ cot x}

.

Substituting

u = cos θ - sin θ cot x, d u = sin θ csc^{2} x d x

I_{2} = \frac{1}{sin θ} \int u^{- 1/2} d u = \frac{2 u}{sin θ} = 2 csc θ cos θ - sin θ cot x

.

So

B = 2 csc θ

A B = 4 sec θ csc θ = \frac{4}{sin θ cos θ} = \frac{8}{sin 2 θ} = 8 csc (2 θ)

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option