`int_(0)^(pi//2)(cos x)/(1+sin^(2)x)dx=`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `(3)/(2)`
D. 0

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin^2 x} \, dx, \]

we will use a substitution method.

### Step 1: Substitution
Let \( t = \sin x \). Then, we differentiate both sides:

\[ dt = \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos x}. \]

### Step 2: Change of Limits
When \( x = 0 \), \( t = \sin(0) = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \sin\left(\frac{\pi}{2}\right) = 1 \).
Thus, the limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \) to \( t = 0 \) to \( t = 1 \).

### Step 3: Rewrite the Integral
Now, substituting \( t \) into the integral gives:

\[ I = \int_{0}^{1} \frac{1}{1 + t^2} \, dt. \]

### Step 4: Integrate
The integral \( \int \frac{1}{1 + t^2} \, dt \) is a standard integral that equals \( \tan^{-1}(t) \). Therefore,

\[ I = \left[ \tan^{-1}(t) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0). \]

### Step 5: Evaluate the Limits
We know that:

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

Thus,

\[ I = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Answer
The value of the integral is

\[ \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin^2 x} \, dx = \frac{\pi}{4}. \]

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