Key Concepts and Formulas

• Trigonometric identities: $\sin(x - \theta) = \sin x \cos \theta - \cos x \sin \theta$, $\tan x = \frac{\sin x}{\cos x}$, $\cot x = \frac{\cos x}{\sin x}$, $\sec x = \frac{1}{\cos x}$, $\csc x = \frac{1}{\sin x}$, $\csc 2\theta = \frac{1}{2\sin \theta \cos \theta}$.
• Indefinite integral properties: $\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx$.
• Integration using substitution: $\int f(g(x))g'(x) dx = \int f(u) du$, where $u = g(x)$.

Step-by-Step Solution

Step 1: Rewrite the integral using the sine difference formula.

We begin by expanding the $\sin(x - \theta)$ term using the trigonometric identity $\sin(x - \theta) = \sin x \cos \theta - \cos x \sin \theta$. $I = \int \frac{\sin^{\frac{3}{2}}x + \cos^{\frac{3}{2}}x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx$

Step 2: Separate the integral into two parts and simplify each.

We separate the numerator into two terms, creating two separate integrals: $I = \int \frac{\sin^{\frac{3}{2}}x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx + \int \frac{\cos^{\frac{3}{2}}x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx$

Now, we simplify the first integral by dividing both numerator and denominator by $\sin^{\frac{3}{2}}x$: $I_1 = \int \frac{\sin^{\frac{3}{2}}x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx = \int \frac{1}{\sqrt{\cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx = \int \frac{1}{\sqrt{\cos^4 x (\tan x \cos \theta - \sin \theta)}} dx = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} dx$

Similarly, we simplify the second integral by dividing both numerator and denominator by $\cos^{\frac{3}{2}}x$: $I_2 = \int \frac{\cos^{\frac{3}{2}}x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx = \int \frac{1}{\sqrt{\sin^3 x (\sin x \cos \theta - \cos x \sin \theta)}} dx = \int \frac{1}{\sqrt{\sin^4 x (\cos \theta - \cot x \sin \theta)}} dx = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} dx$

Therefore, $I = I_1 + I_2$, where $I_1 = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} dx$ $I_2 = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} dx$

Step 3: Evaluate the first integral $I_1$ using substitution.

Let $u = \tan x \cos \theta - \sin \theta$. Then $\frac{du}{dx} = \sec^2 x \cos \theta$, so $du = \sec^2 x \cos \theta \, dx$, which implies $\sec^2 x \, dx = \frac{du}{\cos \theta} = \sec \theta \, du$. $I_1 = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} dx = \int \frac{\sec \theta}{\sqrt{u}} du = \sec \theta \int u^{-\frac{1}{2}} du = \sec \theta \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = 2 \sec \theta \sqrt{u} + C_1 = 2 \sec \theta \sqrt{\tan x \cos \theta - \sin \theta} + C_1$

Step 4: Evaluate the second integral $I_2$ using substitution.

Let $v = \cos \theta - \cot x \sin \theta$. Then $\frac{dv}{dx} = \csc^2 x \sin \theta$, so $dv = \csc^2 x \sin \theta \, dx$, which implies $\csc^2 x \, dx = \frac{dv}{\sin \theta} = \csc \theta \, dv$. $I_2 = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} dx = \int \frac{\csc \theta}{\sqrt{v}} dv = \csc \theta \int v^{-\frac{1}{2}} dv = \csc \theta \cdot \frac{v^{\frac{1}{2}}}{\frac{1}{2}} + C_2 = 2 \csc \theta \sqrt{v} + C_2 = 2 \csc \theta \sqrt{\cos \theta - \cot x \sin \theta} + C_2$

Step 5: Combine the results and compare with the given form.

We have $I = I_1 + I_2$, so $I = 2 \sec \theta \sqrt{\tan x \cos \theta - \sin \theta} + 2 \csc \theta \sqrt{\cos \theta - \cot x \sin \theta} + C$ where $C = C_1 + C_2$.

Comparing this with the given form $I = A \sqrt{\cos \theta \tan x - \sin \theta} + B \sqrt{\cos \theta - \sin \theta \cot x} + C$ we identify $A = 2 \sec \theta$ and $B = 2 \csc \theta$.

Step 6: Calculate the product $AB$.

Then $AB = (2 \sec \theta)(2 \csc \theta) = 4 \sec \theta \csc \theta = 4 \cdot \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{4}{\sin \theta \cos \theta} = \frac{8}{2 \sin \theta \cos \theta} = \frac{8}{\sin 2\theta} = 8 \csc (2\theta)$.

Common Mistakes & Tips

• Be careful with the trigonometric identities, especially when manipulating the expression under the square root.
• Remember to substitute back the original variable after integration.
• Double-check the algebra and arithmetic, especially when calculating $AB$.

Summary

We expanded the given integral using trigonometric identities, split it into two integrals, solved each integral using substitution, and then combined the results. Finally, we compared the result with the given form to find the values of $A$ and $B$, and calculated their product. The product $AB$ simplifies to $8 \csc(2\theta)$.

Final Answer

The final answer is \boxed{8 \operatorname{cosec}(2 \theta)}, which corresponds to option (B).