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

Options

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

Explanation

To solve the integral

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

we will use the substitution \( \sin x = t \).

### Step 1: Substitution
Let \( \sin x = t \). Then, the differential \( dx \) can be expressed as:
\[ \cos x \, dx = dt. \]
This means that \( dx = \frac{dt}{\cos x} \).

### Step 2: Change of Limits
Next, we need to change the limits of integration. When \( x = 0 \):
\[ t = \sin(0) = 0. \]
When \( x = \frac{\pi}{2} \):
\[ t = \sin\left(\frac{\pi}{2}\right) = 1. \]
Thus, the new limits of integration are from \( t = 0 \) to \( t = 1 \).

### Step 3: Rewrite the Integral
Substituting into the integral, we have:
\[ I = \int_{0}^{1} \frac{dt}{1 + t^2}. \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{dt}{1 + t^2} \) is a standard integral, and its result is:
\[ \int \frac{dt}{1 + t^2} = \tan^{-1}(t). \]
Thus, we can evaluate:
\[ I = \left[ \tan^{-1}(t) \right]_{0}^{1}. \]

### Step 5: Apply the Limits
Now we apply the limits:
\[ I = \tan^{-1}(1) - \tan^{-1}(0). \]
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \quad \text{and} \quad \tan^{-1}(0) = 0. \]
Therefore,
\[ I = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Answer
Thus, 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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