Evaluate : `int_(pi/5)^((3pi)/(10))(cosx)/(sinx+cosx)dx `

Answer

To evaluate the integral

\[ I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{\cos x}{\sin x + \cos x} \, dx, \]

we will use the property of definite integrals:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

### Step 1: Apply the property of definite integrals

Let \( a = \frac{\pi}{5} \) and \( b = \frac{3\pi}{10} \). First, we calculate \( a + b \):

\[ a + b = \frac{\pi}{5} + \frac{3\pi}{10} = \frac{2\pi}{10} + \frac{3\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}. \]

Now, we can rewrite the integral using the property:

\[ I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{\cos\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx. \]

### Step 2: Simplify the integrand

Using the trigonometric identities:

\[ \cos\left(\frac{\pi}{2} - x\right) = \sin x, \]
\[ \sin\left(\frac{\pi}{2} - x\right) = \cos x, \]

we can rewrite the integral:

\[ I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{\sin x}{\cos x + \sin x} \, dx. \]

### Step 3: Add the two integrals

Now we have two expressions for \( I \):

1. \( I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{\cos x}{\sin x + \cos x} \, dx \)
2. \( I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{\sin x}{\sin x + \cos x} \, dx \)

Adding these two equations:

\[ 2I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \left( \frac{\cos x + \sin x}{\sin x + \cos x} \right) \, dx = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} 1 \, dx. \]

### Step 4: Evaluate the integral of 1

The integral of 1 over the limits \( \frac{\pi}{5} \) to \( \frac{3\pi}{10} \) is:

\[ \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} 1 \, dx = \left[ x \right]_{\frac{\pi}{5}}^{\frac{3\pi}{10}} = \frac{3\pi}{10} - \frac{\pi}{5}. \]

Calculating this:

\[ \frac{3\pi}{10} - \frac{2\pi}{10} = \frac{\pi}{10}. \]

### Step 5: Solve for \( I \)

Now we have:

\[ 2I = \frac{\pi}{10} \implies I = \frac{\pi}{20}. \]

Thus, the final answer is:

\[ \boxed{\frac{\pi}{20}}. \]