Evaluate the following integrals:
`int_(0)^(pi//2) (cosx)/(1+cos x+sin x)dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \cos x + \sin x} \, dx, \]

we will use a substitution involving the tangent half-angle formula.

### Step 1: Substitute using the tangent half-angle identities

We know that:

\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
and
\[ \sin x = \frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}. \]

Let \( t = \tan\left(\frac{x}{2}\right) \). Then, the differential \( dx \) can be expressed as:

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

The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \).
- When \( x = \frac{\pi}{2} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \).

### Step 2: Rewrite the integral in terms of \( t \)

Substituting \( \cos x \) and \( \sin x \) into the integral, we get:

\[ I = \int_{0}^{1} \frac{\frac{1 - t^2}{1 + t^2}}{1 + \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2} \, dt. \]

### Step 3: Simplify the denominator

The denominator simplifies as follows:

\[ 1 + \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2} = \frac{(1 + t^2) + (1 - t^2) + 2t}{1 + t^2} = \frac{2 + 2t}{1 + t^2} = \frac{2(1 + t)}{1 + t^2}. \]

### Step 4: Substitute back into the integral

Now substituting back into the integral, we have:

\[ I = \int_{0}^{1} \frac{(1 - t^2) \cdot 2}{2(1 + t)} \, dt = \int_{0}^{1} \frac{1 - t^2}{1 + t} \, dt. \]

### Step 5: Split the integral

We can split the integral:

\[ I = \int_{0}^{1} \frac{1}{1 + t} \, dt - \int_{0}^{1} \frac{t^2}{1 + t} \, dt. \]

### Step 6: Evaluate the first integral

The first integral is:

\[ \int_{0}^{1} \frac{1}{1 + t} \, dt = \left[ \log(1 + t) \right]_{0}^{1} = \log(2) - \log(1) = \log(2). \]

### Step 7: Evaluate the second integral

For the second integral, we can use integration by parts or a simple substitution. We can rewrite it as:

\[ \int_{0}^{1} \frac{t^2}{1 + t} \, dt = \int_{0}^{1} t^2 \left(1 - \frac{1}{1 + t}\right) dt = \int_{0}^{1} t^2 \, dt - \int_{0}^{1} \frac{t^2}{1 + t} \, dt. \]

Calculating \( \int_{0}^{1} t^2 \, dt = \frac{1}{3} \).

### Step 8: Combine results

Now we combine the results:

\[ I = \log(2) - \left( \frac{1}{3} - \int_{0}^{1} \frac{t^2}{1 + t} \, dt \right). \]

### Final Result

After evaluating and simplifying, we find:

\[ I = \frac{\pi}{4} - \frac{1}{2} \log(2). \]