The value of `(dI)/(da)` when `I=int_(0)^(pi//2) log((1+asinx)/(1-asinx)) (dx)/(sinx)` (where `|a|lt1`) is

Options

A. `(pi)/(sqrt(1-a^(2)))`
B. `-pisqrt(1-a^(2))`
C. `sqrt(1-a^(2))`
D. `(sqrt(1-a^(2)))/(pi)`

Explanation

To find the value of \(\frac{dI}{da}\) where

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\log\left(\frac{1 + a \sin x}{1 - a \sin x}\right)}{\sin x} \, dx \]

for \(|a| < 1\), we will proceed step by step.

### Step 1: Differentiate under the integral sign
We start by differentiating \(I\) with respect to \(a\):

\[ \frac{dI}{da} = \int_{0}^{\frac{\pi}{2}} \frac{\partial}{\partial a} \left( \frac{\log\left(\frac{1 + a \sin x}{1 - a \sin x}\right)}{\sin x} \right) dx \]

### Step 2: Differentiate the logarithm
Using the chain rule, we differentiate the logarithm:

\[ \frac{\partial}{\partial a} \log\left(\frac{1 + a \sin x}{1 - a \sin x}\right) = \frac{\sin x}{1 + a \sin x} + \frac{\sin x}{1 - a \sin x} \]

Thus, we have:

\[ \frac{dI}{da} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x} \left( \frac{\sin x}{1 + a \sin x} + \frac{\sin x}{1 - a \sin x} \right) dx \]

### Step 3: Simplify the expression
This simplifies to:

\[ \frac{dI}{da} = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{1 + a \sin x} + \frac{1}{1 - a \sin x} \right) dx \]

### Step 4: Combine the fractions
Now, we combine the fractions:

\[ \frac{1}{1 + a \sin x} + \frac{1}{1 - a \sin x} = \frac{(1 - a \sin x) + (1 + a \sin x)}{(1 + a \sin x)(1 - a \sin x)} = \frac{2}{1 - a^2 \sin^2 x} \]

### Step 5: Substitute back into the integral
Substituting this back into our integral gives:

\[ \frac{dI}{da} = \int_{0}^{\frac{\pi}{2}} \frac{2}{1 - a^2 \sin^2 x} \, dx \]

### Step 6: Evaluate the integral
The integral can be evaluated using the known result:

\[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 - k^2 \sin^2 x} = \frac{\pi}{2\sqrt{1-k^2}} \quad \text{for } |k| < 1 \]

In our case, \(k^2 = a^2\), so we have:

\[ \int_{0}^{\frac{\pi}{2}} \frac{2}{1 - a^2 \sin^2 x} \, dx = 2 \cdot \frac{\pi}{2\sqrt{1-a^2}} = \frac{\pi}{\sqrt{1-a^2}} \]

### Final Result
Thus, we find:

\[ \frac{dI}{da} = \frac{\pi}{\sqrt{1-a^2}} \]