Evaluate $\int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} dx$.

$\sin^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} + \log \left| \sqrt{1-x^2} \right| + C$

The correct answer is Option (1) → $\sin^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} + \log \left| \sqrt{1-x^2} \right| + C$

Let $I = \int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} dx = \int \frac{\sin^{-1} x}{(1 - x^2) \sqrt{1 - x^2}} dx$

Let $\sin^{-1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt$

Now $I = \int \frac{t}{(1 - x^2)} dt$

$∵\sin^{-1} x = t \Rightarrow x = \sin t \Rightarrow 1 - x^2 = 1 - \sin^2 t = \cos^2 t \text{}$

$= \int \frac{t}{\cos^2 t} dt = \int t \sec^2 t \, dt$

$= t \int \sec^2 t \, dt - \int \left[ \frac{d}{dt} t \int (\sec^2 t \, dt) \right] dt \quad \text{[Integration by parts]} \text{}$

$= t \cdot \tan t - \int 1 \cdot \tan t \, dt$

$= t \cdot \tan t + \log |\cos t| + C \quad \left[ ∵\int \tan x \, dx = -\log |\cos x| + C \right]$

$= t \left( \frac{\sin t}{\cos t} \right) + \log |\cos t| + C \quad \left[ ∵\tan t = \frac{\sin t}{\cos t} \right]$

$= \sin^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} + \log \left| \sqrt{1-x^2} \right| + C$