Choose the correct Answer of the Following Questions : `int_0^(pi/2) (sinx dx)/(sinx + cosx)=`

Options

A. `pi`
B. `pi/2`
C. `pi/4`
D. `pi/6`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx \), we will use a symmetry property of definite integrals. Here’s the step-by-step solution:

### Step 1: Set up the integral
Let
\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx \]

### Step 2: Use the substitution \( x = \frac{\pi}{2} - x \)
We will apply the substitution \( x = \frac{\pi}{2} - t \), which implies \( dx = -dt \). When \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \). Therefore, we have:
\[ I = \int_{\frac{\pi}{2}}^0 \frac{\sin\left(\frac{\pi}{2} - t\right)}{\sin\left(\frac{\pi}{2} - t\right) + \cos\left(\frac{\pi}{2} - t\right)} (-dt) \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos t}{\cos t + \sin t} \, dt \]

### Step 3: Rewrite the integral
Now, we can express \( I \) in terms of the new variable \( t \):
\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx \]

### Step 4: Add the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx \)
2. \( I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx \)

Adding these two integrals:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\sin x + \cos x} \right) dx \]

### Step 5: Simplify the integrand
The integrand simplifies to:
\[ \frac{\sin x + \cos x}{\sin x + \cos x} = 1 \]
Thus, we have:
\[ 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \]

### 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 \( I \)
Dividing both sides by 2 gives:
\[ I = \frac{\pi}{4} \]

### Conclusion
Thus, the value of the integral is:
\[ \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx = \frac{\pi}{4} \]