The value of
`int_(0)^(pi//2)(log(1+x sin^(2) theta))/(sin^(2)theta)d theta,xge0` is equal to

Options

A. `pi(sqrt(1+x-1))`
B. `pi(sqrt(1+x-2))`
C. `sqrtpi(sqrt(1+x-1))`
D. None of these

Answer

To solve the integral

\[ I(x) = \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + x \sin^2 \theta)}{\sin^2 \theta} \, d\theta \]

where \( x \geq 0 \), we will follow these steps:

### Step 1: Evaluate \( I(0) \)

First, we substitute \( x = 0 \) into the integral:

\[ I(0) = \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + 0 \cdot \sin^2 \theta)}{\sin^2 \theta} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\log(1)}{\sin^2 \theta} \, d\theta = \int_{0}^{\frac{\pi}{2}} 0 \, d\theta = 0 \]

### Step 2: Differentiate \( I(x) \) with respect to \( x \)

Next, we differentiate \( I(x) \):

\[ I'(x) = \frac{d}{dx} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + x \sin^2 \theta)}{\sin^2 \theta} \, d\theta \]

Using Leibniz's rule for differentiation under the integral sign, we have:

\[ I'(x) = \int_{0}^{\frac{\pi}{2}} \frac{\partial}{\partial x} \left( \frac{\log(1 + x \sin^2 \theta)}{\sin^2 \theta} \right) d\theta \]

Calculating the derivative inside the integral:

\[ \frac{\partial}{\partial x} \log(1 + x \sin^2 \theta) = \frac{\sin^2 \theta}{1 + x \sin^2 \theta} \]

Thus,

\[ I'(x) = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 \theta}{1 + x \sin^2 \theta} \, d\theta \]

### Step 3: Solve \( I'(x) \)

To solve \( I'(x) \), we can use the substitution \( u = \sin^2 \theta \), which gives \( du = 2 \sin \theta \cos \theta \, d\theta \) or \( d\theta = \frac{du}{2\sqrt{u(1-u)}} \). The limits change from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) to \( u = 0 \) to \( u = 1 \):

\[ I'(x) = \int_{0}^{1} \frac{u}{1 + xu} \cdot \frac{1}{2\sqrt{u(1-u)}} \, du \]

This integral can be solved using standard techniques or known results.

### Step 4: Integrate \( I'(x) \)

The integral can be simplified and evaluated, leading to:

\[ I'(x) = \frac{\pi}{2\sqrt{x(1+x)}} \]

### Step 5: Integrate \( I'(x) \) to find \( I(x) \)

Now, we integrate \( I'(x) \):

\[ I(x) = \int I'(x) \, dx = \int \frac{\pi}{2\sqrt{x(1+x)}} \, dx \]

Using the substitution \( u = \sqrt{x(1+x)} \), we can evaluate this integral. The result will be:

\[ I(x) = \frac{\pi}{2} (\sqrt{1+x} - 1) + C \]

### Step 6: Determine the constant \( C \)

From our earlier evaluation, we know \( I(0) = 0 \):

\[ I(0) = \frac{\pi}{2} (1 - 1) + C = 0 \implies C = 0 \]

Thus, the final result is:

\[ I(x) = \frac{\pi}{2} (\sqrt{1+x} - 1) \]

### Final Answer

The value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + x \sin^2 \theta)}{\sin^2 \theta} \, d\theta = \frac{\pi}{2} (\sqrt{1+x} - 1) \]