The value of $\frac{dI}{da}$ when $I = \int_0^{\pi/2} \log\left(\frac{1 + a \sin x}{1 - a \sin x}\right) \frac{dx}{\sin x}$ (where $|a| < 1$) is:
Step-by-Step Solution
Key Concept: Differentiating under the integral sign with respect to parameter $a$ simplifies the integrand, then integrate back to recover the original function.
Let $I(a) = \int_0^{\pi/2} \log\left(\frac{1+a\sin x}{(1-a\sin x)\sin x}\right) dx$. Differentiating with respect to $a$ gives $\frac{dI}{da} = \int_0^{\pi/2} \frac{2\sin x}{1-a^2\sin^2 x} dx = \int_0^{\pi/2} \frac{2\sec^2 x dx}{1+(1-a^2)\tan^2 x}$. Using substitution $t = \tan x$ and evaluating yields $\frac{dI}{da} = \frac{2}{\sqrt{1-a^2}}\tan^{-1}(t\sqrt{1-a^2})|_0^{\pi/2} = \frac{\pi}{\sqrt{1-a^2}}$. Integrating: $I(a) = \pi\sin^{-1}(a) + c$. With $I(0) = 0$, we get $c = 0$, so $I = \pi\sin^{-1}(a)$.
Correct Answer: 1