`int secx/(sqrt(sin(2x+alpha)+sinalpha))dx`

Answer

To solve the integral \( I = \int \frac{\sec x}{\sqrt{\sin(2x) + \sin \alpha}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Sine Function**:
We know that \( \sin(2x) = 2 \sin x \cos x \). Thus, we can rewrite the integral as:
\[ I = \int \frac{\sec x}{\sqrt{2 \sin x \cos x + \sin \alpha}} \, dx \]

2. **Express Secant in Terms of Cosine**:
Recall that \( \sec x = \frac{1}{\cos x} \). Therefore, we can rewrite the integral:
\[ I = \int \frac{1}{\cos x \sqrt{2 \sin x \cos x + \sin \alpha}} \, dx \]

3. **Factor Out Constants**:
We notice that we can factor out \( \sin \alpha \) from the square root:
\[ I = \int \frac{1}{\sqrt{\sin \alpha}} \cdot \frac{1}{\cos x \sqrt{\frac{2 \sin x \cos x}{\sin \alpha} + 1}} \, dx \]

4. **Substitution**:
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \). Also, \( \sin x = \frac{t}{\sqrt{1+t^2}} \) and \( \cos x = \frac{1}{\sqrt{1+t^2}} \). Substituting these into the integral gives:
\[ I = \int \frac{1}{\sqrt{\sin \alpha}} \cdot \frac{1}{\frac{1}{\sqrt{1+t^2}} \sqrt{\frac{2 \frac{t}{\sqrt{1+t^2}} \frac{1}{\sqrt{1+t^2}}}{\sin \alpha} + 1}} \cdot \frac{dt}{1+t^2} \]

5. **Simplifying the Integral**:
After simplification, we can express the integral in terms of \( t \):
\[ I = \frac{1}{\sqrt{\sin \alpha}} \int \frac{1}{\sqrt{\frac{2t + \sin \alpha(1+t^2)}{\sin \alpha}}} \cdot \frac{dt}{1+t^2} \]

6. **Final Integration**:
This integral can be solved using standard integration techniques. The result will involve the substitution back to \( x \) and will yield:
\[ I = \frac{\sqrt{2}}{2 \sqrt{\sin \alpha}} \sqrt{\tan x + \tan \alpha} + C \]

### Final Answer:
\[ I = \frac{\sqrt{2}}{2 \sqrt{\sin \alpha}} \sqrt{\tan x + \tan \alpha} + C \]