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

Expand $$\sin(x-\theta) = \sin x \cos \theta - \cos x \sin \theta$$.

Divide the numerator terms individually by the denominator:

$$I = \int \frac{\sin^{3/2}x}{\sqrt{\sin^3 x \cos^3 x \sin(x-\theta)}} dx + \int \frac{\cos^{3/2}x}{\sqrt{\sin^3 x \cos^3 x \sin(x-\theta)}} dx$$

$$I = \int \frac{1}{\sqrt{\cos^3 x \sin(x-\theta)}} dx + \int \frac{1}{\sqrt{\sin^3 x \sin(x-\theta)}} dx$$

For the first part $$I_1$$:

$$I_1 = \int \frac{1}{\cos^2 x \sqrt{\frac{\sin(x-\theta)}{\cos x}}} dx = \int \frac{\sec^2 x}{\sqrt{\frac{\sin x \cos \theta - \cos x \sin \theta}{\cos x}}} dx = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} dx$$

Let $$u = \tan x \cos \theta - \sin \theta$$, then $$du = \sec^2 x \cos \theta dx$$.

$$I_1 = \frac{1}{\cos \theta} \int u^{-1/2} du = \frac{2}{\cos \theta} \sqrt{\tan x \cos \theta - \sin \theta}$$

For the second part $$I_2$$:

$$I_2 = \int \frac{1}{\sin^2 x \sqrt{\frac{\sin(x-\theta)}{\sin x}}} dx = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} dx$$

Let $$v = \cos \theta - \cot x \sin \theta$$, then $$dv = \csc^2 x \sin \theta dx$$.

$$I_2 = \frac{1}{\sin \theta} \int v^{-1/2} dv = \frac{2}{\sin \theta} \sqrt{\cos \theta - \cot x \sin \theta}$$

Comparing with the given form $$A\sqrt{\dots} + B\sqrt{\dots}$$:

• $$A = \frac{2}{\cos \theta}$$

• $$B = \frac{2}{\sin \theta}$$

$$AB = \left( \frac{2}{\cos \theta} \right) \left( \frac{2}{\sin \theta} \right) = \frac{4}{\sin \theta \cos \theta}$$

$$AB = \frac{8}{2 \sin \theta \cos \theta} = \frac{8}{\sin(2\theta)} = \mathbf{8 \csc(2\theta)}$$