`int_(0)^((pi)/2)(cos2x)/(cosx+sinx)dx=`

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To solve the integral

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

we will use a property of definite integrals. The property states that if \( I = \int_{a}^{b} f(x) \, dx \), then

\[ I = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, we have \( a = 0 \) and \( b = \frac{\pi}{2} \). Thus, we can rewrite the integral as follows:

1. **Step 1: Apply the property**

We apply the property to our integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos 2\left(\frac{\pi}{2} - x\right)}{\cos\left(\frac{\pi}{2} - x\right) + \sin\left(\frac{\pi}{2} - x\right)} \, dx. \]

2. **Step 2: Simplify the expression**

Now, we simplify the terms inside the integral:

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

Therefore, we can rewrite the integral as:

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

3. **Step 3: Combine the two integrals**

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

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

Adding these two equations together:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos 2x}{\cos x + \sin x} - \frac{\cos 2x}{\cos x + \sin x} \right) dx = 0. \]

4. **Step 4: Solve for \( I \)**

From the equation \( 2I = 0 \), we can conclude that:

\[ I = 0. \]

Thus, the value of the integral is

\[ \boxed{0}. \]