If sin 3 2 x + cos 3 2 x sin 3 x cos 3 x sin ( x - ) d x = A cos tan x - sin + B cos - sin cot x + C , where C is the integration constant, then AB is equal to [2024]

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Solution

Q.
If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

1 $4 c o s e c (2 θ)$

2 $2 \sec θ$

3 $4 \sec θ$

4 $8 c o s e c (2 θ)$

Ans.
(4)

Let $I = \int \frac{\sin^{3 / 2} x + \cos^{3 / 2} x}{\sqrt{\sin^{3} x \cdot \cos^{3} x \cdot \sin (x - θ)}} d x$

$= \int \frac{d x}{\sqrt{\cos^{3} x \cdot \sin (x - θ)}} + \int \frac{d x}{\sqrt{\sin^{3} x \cdot \sin (x - θ)}}$

$= \int \frac{d x}{\cos^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\cos x}}} + \int \frac{d x}{\sin^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\sin x}}}$

$= \int \frac{\sec^{2} x}{\sqrt{\tan x \cos θ - \sin θ}} d x + \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

$\Rightarrow I = I_{1} + I_{2} (Let)$ ...(i)

Now, $I_{1} = \int \frac{\sec^{2} x d x}{\sqrt{\tan x \cos θ - \sin θ}}$

Put $\tan x \cos θ - \sin θ = t \Rightarrow \cos θ \cdot \sec^{2} x d x = d t$

$∴ I_{1} = \int \frac{d t}{\cos θ \cdot \sqrt{t}} \Rightarrow I_{1} = \frac{2 t^{1 / 2}}{\cos θ}$

$\Rightarrow I_{1} = \frac{2 {(\tan x \cos θ - \sin θ)}^{1 / 2}}{\cos θ}$

And, $I_{2} = \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

Put $\cos θ - \cot x \sin θ = v \Rightarrow \sin θ c o s e c^{2} x d x = d v$

$∴ I_{2} = \int \frac{d v}{\sin θ \sqrt{v}} = \frac{2 \sqrt{v}}{\sin θ} \Rightarrow I_{2} = \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ}$

From (i),

$I = \frac{2 \sqrt{\tan x \cos θ - \sin θ}}{\cos θ} + \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ} + C$ and

$I = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C$

So, $A = \frac{2}{\cos θ}, B = \frac{2}{\sin θ}$

$∴ A B = \frac{4 \cdot 2}{2 \sin θ \cos θ} = 8 c o s e c 2 θ$

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