Evaluate the following integrals:
`int(1+sinx)/(sin x(1+cosx))dx`

Answer

To evaluate the integral

\[ I = \int \frac{1 + \sin x}{\sin x (1 + \cos x)} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form:

\[ I = \int \frac{1 + \sin x}{\sin x (1 + \cos x)} \, dx = \int \frac{1}{\sin x (1 + \cos x)} \, dx + \int \frac{\sin x}{\sin x (1 + \cos x)} \, dx. \]

This simplifies to:

\[ I = \int \frac{1}{\sin x (1 + \cos x)} \, dx + \int \frac{1}{1 + \cos x} \, dx. \]

### Step 2: Simplify the First Integral
For the first integral, we can use the identity \( \sin x = 2 \tan \frac{x}{2} \frac{1}{1 + \tan^2 \frac{x}{2}} \) and \( \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \).

Let \( t = \tan \frac{x}{2} \). Then, we have:

\[ \sin x = \frac{2t}{1 + t^2} \quad \text{and} \quad 1 + \cos x = \frac{2}{1 + t^2}. \]

Substituting these into the integral:

\[ \int \frac{1}{\sin x (1 + \cos x)} \, dx = \int \frac{1 + t^2}{2t} \cdot \frac{1 + t^2}{2} \, dx. \]

### Step 3: Change of Variable
Using the substitution \( t = \tan \frac{x}{2} \), we differentiate to find \( dx \):

\[ dx = \frac{2}{1 + t^2} dt. \]

Substituting \( dx \) into the integral gives:

\[ \int \frac{1 + t^2}{2t} \cdot \frac{1 + t^2}{2} \cdot \frac{2}{1 + t^2} dt = \int \frac{1 + t^2}{2t} dt. \]

### Step 4: Split the Integral
Now we can split the integral:

\[ \int \frac{1 + t^2}{2t} dt = \frac{1}{2} \int \frac{1}{t} dt + \frac{1}{2} \int t \, dt. \]

### Step 5: Evaluate the Integrals
Evaluating these integrals gives:

\[ \frac{1}{2} \log |t| + \frac{1}{4} t^2 + C. \]

### Step 6: Substitute Back
Substituting back \( t = \tan \frac{x}{2} \):

\[ I = \frac{1}{2} \log \left| \tan \frac{x}{2} \right| + \frac{1}{4} \tan^2 \frac{x}{2} + C. \]

### Final Answer
Thus, the final answer for the integral is:

\[ I = \frac{1}{2} \log \left| \tan \frac{x}{2} \right| + \frac{1}{4} \tan^2 \frac{x}{2} + C. \]