`int_(0)^(pi//2) ( f ( co x )) /( f (sinx ) + f(co sx) ) dx` is equal to

Options

A. 0
B. `(pi)/(2)`
C. `(pi)/( 4)`
D. none of these

Answer

To solve the integral

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

we will use a property of definite integrals.

### Step 1: Define the Integral
Let

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

### Step 2: Use the Integral Property
We can use the property of integrals that states

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

In our case, we will substitute \( x \) with \( \frac{\pi}{2} - x \):

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

### Step 3: Simplify the Integral
Now, we know that

\[ \cos\left(\frac{\pi}{2} - x\right) = \sin x \quad \text{and} \quad \sin\left(\frac{\pi}{2} - x\right) = \cos x. \]

Thus, we can rewrite the integral as:

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

### Step 4: Add the Two Integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx \)

Adding these two equations gives:

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

### Step 5: Simplify Further
The expression simplifies to:

\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx. \]

### Step 6: Evaluate the Integral
Now we can evaluate the integral:

\[ 2I = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Step 7: Solve for \( I \)
Now, we can solve for \( I \):

\[ I = \frac{\pi}{4}. \]

### Conclusion
Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]