`int((sinx+cosx))/(sqrt(sin2x))dx`

Answer

To solve the integral \( I = \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx \]
Recall that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx \]

### Step 2: Simplify the Denominator
The square root in the denominator can be simplified:
\[ \sqrt{2 \sin x \cos x} = \sqrt{2} \sqrt{\sin x \cos x} \]
Thus, we have:
\[ I = \int \frac{\sin x + \cos x}{\sqrt{2} \sqrt{\sin x \cos x}} \, dx \]
This simplifies to:
\[ I = \frac{1}{\sqrt{2}} \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx \]

### Step 3: Use Trigonometric Identities
We can express \( \sin x + \cos x \) in terms of \( \sin x \) and \( \cos x \):
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
Now, substituting this back into the integral:
\[ I = \frac{1}{\sqrt{2}} \int \frac{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}{\sqrt{\sin x \cos x}} \, dx \]
This simplifies to:
\[ I = \int \frac{\sin\left(x + \frac{\pi}{4}\right)}{\sqrt{\sin x \cos x}} \, dx \]

### Step 4: Substitute \( t = \sin x - \cos x \)
Let \( t = \sin x - \cos x \). Then, differentiating gives:
\[ dt = (\cos x + \sin x) \, dx \]
Thus, we can express \( dx \) as:
\[ dx = \frac{dt}{\cos x + \sin x} \]

### Step 5: Substitute Back into the Integral
Now substituting \( t \) into the integral:
\[ I = \int \frac{dt}{\sqrt{1 - t^2}} \]
This integral is known and evaluates to:
\[ I = \sin^{-1}(t) + C \]

### Step 6: Substitute Back for \( t \)
Substituting back for \( t \):
\[ I = \sin^{-1}(\sin x - \cos x) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \sin^{-1}(\sin x - \cos x) + C \]