`int(1)/(sqrt(sin^(3)xsin(x+alpha)))dx` is equal to

Options

A. `2"cosec "alphasqrt(cosalpha+sin alphatanx)+C`
B. `-2"cosec " alphasqrt(cos alpha+sinalphacotx)+C`
C. `"cosec "alphasqrt(cosalpha+sinalphacotx)+C`
D. none of these

Answer

To solve the integral \( I = \int \frac{1}{\sqrt{\sin^3 x \sin(x + \alpha)}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \frac{1}{\sqrt{\sin^3 x \sin(x + \alpha)}} \, dx \]

2. **Use the Sine Addition Formula**:
The sine addition formula states:
\[ \sin(x + \alpha) = \sin x \cos \alpha + \cos x \sin \alpha \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{1}{\sqrt{\sin^3 x \left(\sin x \cos \alpha + \cos x \sin \alpha\right)}} \, dx \]

3. **Factor Out \(\sin^3 x\)**:
We can factor out \(\sin^3 x\) from the square root:
\[ I = \int \frac{1}{\sqrt{\sin^4 x \cos \alpha + \sin^3 x \cos x \sin \alpha}} \, dx \]

4. **Substitution**:
Let \( t = \cos \alpha + \cot x \sin \alpha \). Then, differentiate \( t \):
\[ dt = -\csc^2 x \sin \alpha \, dx \implies dx = -\frac{dt}{\sin \alpha \csc^2 x} \]

5. **Express \( \csc^2 x \)**:
Since \( \csc^2 x = \frac{1}{\sin^2 x} \), we can substitute:
\[ dx = -\frac{\sin^2 x}{\sin \alpha} dt \]

6. **Substituting in the Integral**:
Substitute \( dx \) into the integral:
\[ I = \int \frac{-\sin^2 x}{\sin \alpha \sqrt{t}} \, dt \]

7. **Simplify the Integral**:
The integral simplifies to:
\[ I = -\frac{1}{\sin \alpha} \int \frac{1}{\sqrt{t}} \, dt \]

8. **Integrate**:
The integral of \( \frac{1}{\sqrt{t}} \) is:
\[ \int t^{-1/2} \, dt = 2t^{1/2} + C \]
Therefore:
\[ I = -\frac{2}{\sin \alpha} \sqrt{t} + C \]

9. **Substituting Back for \( t \)**:
Substitute back \( t = \cos \alpha + \cot x \sin \alpha \):
\[ I = -\frac{2}{\sin \alpha} \sqrt{\cos \alpha + \cot x \sin \alpha} + C \]

10. **Final Result**:
Thus, the final answer is:
\[ I = -\frac{2}{\sin \alpha} \sqrt{\cos \alpha + \cot x \sin \alpha} + C \]