Evaluate the integrals`int_0^(pi/2)(sinx)/(1+cos^2x)dx`

Options

A. `pi/4`
B. 0
C. `pi/2`
D. None of these

Explanation

To evaluate the integral

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} \, dx, \]

we will use the substitution method.

### Step 1: Substitution
Let \( u = \cos x \). Then, the differential \( du = -\sin x \, dx \).

### Step 2: Change of Limits
When \( x = 0 \), \( u = \cos(0) = 1 \).
When \( x = \frac{\pi}{2} \), \( u = \cos\left(\frac{\pi}{2}\right) = 0 \).

Thus, the limits change from \( x: 0 \to \frac{\pi}{2} \) to \( u: 1 \to 0 \).

### Step 3: Rewrite the Integral
Substituting into the integral, we have:

\[ I = \int_1^0 \frac{-1}{1 + u^2} \, du = -\int_1^0 \frac{1}{1 + u^2} \, du. \]

### Step 4: Change the Limits
Reversing the limits of integration gives:

\[ I = \int_0^1 \frac{1}{1 + u^2} \, du. \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{1}{1 + u^2} \, du \) is known to be \( \tan^{-1}(u) \). Therefore,

\[ I = \left[ \tan^{-1}(u) \right]_0^1. \]

Calculating this, we get:

\[ I = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Answer
Thus, the value of the integral is

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

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