Evaluate the following integrals:
`int (sinx)/(sin2x)dx`

Answer

To evaluate the integral \( \int \frac{\sin x}{\sin 2x} \, dx \), we will follow these steps:

### Step 1: Rewrite \(\sin 2x\)
We know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sin x}{\sin 2x} \, dx = \int \frac{\sin x}{2 \sin x \cos x} \, dx \]

### Step 2: Simplify the Integral
Now, we can simplify the integrand:
\[ \int \frac{\sin x}{2 \sin x \cos x} \, dx = \int \frac{1}{2 \cos x} \, dx \]
The \(\sin x\) in the numerator and denominator cancels out.

### Step 3: Factor out the constant
We can factor out the constant \( \frac{1}{2} \):
\[ \int \frac{1}{2 \cos x} \, dx = \frac{1}{2} \int \sec x \, dx \]

### Step 4: Integrate \(\sec x\)
The integral of \(\sec x\) is known:
\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C \]
Thus, we have:
\[ \frac{1}{2} \int \sec x \, dx = \frac{1}{2} \left( \ln |\sec x + \tan x| + C \right) \]

### Step 5: Write the final answer
Putting it all together, we get:
\[ \int \frac{\sin x}{\sin 2x} \, dx = \frac{1}{2} \ln |\sec x + \tan x| + C \]

### Summary of the Solution
The final result is:
\[ \int \frac{\sin x}{\sin 2x} \, dx = \frac{1}{2} \ln |\sec x + \tan x| + C \]