If ∫ sin(3)/(2)x + cos(3)/(2)x ^ 3 x ^ 3 xsin ( x - θ ) dx = A cosθ...

Let, $I = ∫ \dfrac{sin\dfrac{3}{2}x + cos\dfrac{3}{2}x}{\sqrt{\sin^{3}x\cos^{3}xsin ( x - \theta )}}dx$

$⇒ I = ∫ \dfrac{sin\dfrac{3}{2}x + cos\dfrac{3}{2}x}{\sqrt{\sin^{3}x\cos^{3}x ( \sin x \cos\theta - \cos x \sin\theta )}}dx$

$⇒ I = ∫ \dfrac{sin\dfrac{3}{2}x + cos\dfrac{3}{2}x}{\sqrt{\sin^{3}x\cos^{4}x ( \tan x \cos\theta - sin\theta )}}dx$

$⇒ I = ∫ \dfrac{sin\dfrac{3}{2}x}{sin\dfrac{3}{2}x\cos^{2}x\sqrt{\tan x \cos\theta - sin\theta}}dx + ∫ \dfrac{cos\dfrac{3}{2}x}{\sin^{2}xcos\dfrac{3}{2}x\sqrt{cos\theta - \cot x \sin\theta}}dx$

$⇒ I = ∫ \dfrac{1}{\cos^{2}x\sqrt{\tan x \cos\theta - sin\theta}}dx + ∫ \dfrac{1}{\sin^{2}x\sqrt{cos\theta - \cot x \sin\theta}}dx$

$⇒ I = ∫ \dfrac{\sec^{2}x}{\sqrt{\tan x \cos\theta - sin\theta}}dx + ∫ \dfrac{{cosec}^{2}x}{\sqrt{cos\theta - \cot x \sin\theta}}dx$

So, $I = {I}_{1} + {I}_{2} . . . . \left\{let\right\}$

For ${I}_{1}$, let $\tan x \cos\theta - sin\theta = {t}^{2}$

$⇒ cos\theta \times \sec^{2}xdx = 2tdt$

$⇒ \sec^{2}xdx = \dfrac{2tdt}{cos\theta}$

For ${I}_{2}$, let $cos\theta - \cot x \sin\theta = {z}^{2}$

$⇒ {cosec}^{2}x \times sin\thetadx = 2zdz$

$⇒ {cosec}^{2}xdx = \dfrac{2zdz}{sin\theta}$

$⇒ I = ∫ \dfrac{2tdt}{cos\theta \times t} + ∫ \dfrac{2zdz}{sin\theta \times z}$

$⇒ I = 2\left[\left(sec\theta\right) ∫ \left(1\right)dt + cosec\theta ∫ \left(1\right)dz\right]$

$⇒ I = 2sec\theta\sqrt{\tan x \cos\theta - sin\theta} + 2 cosec \theta\sqrt{cos\theta - \cot x \sin\theta}$

$⇒ A \times B = 4sec\theta cosec\theta$

$⇒ A \times B = \dfrac{2}{2} \times \dfrac{4}{sin\theta cos\theta}$

$⇒ A \times B = 8 cosec 2\theta$