Question

If \(f(x) = \frac{sin^{-1}x}{\sqrt{1-x^2}}\) and \(g(x) = e^{sin^{-1}x}\), then the value of \(\int f(x)g(x)\,dx = \ldots\)

• A. \(e^{sin^{-1}x}(sin^{-1}x-1)+c\)
• B. \(e^{sin^{-1}x}(1-sin^{-1}x)+c\)
• C. \(e^{sin^{-1}x}(sin^{-1}x+1)+c\)
• D. \(e^{sin^{-1}x}(-sin^{-1}x-1)+c\)

Hint:

Substitute x = a tan^2 theta, which makes the inverse cosecant equal to theta, then integrate by parts.

Answer 1

The Correct Option is A

Step 1: Switch to the variable theta:
With $x = a\tan^2\theta$ the integrand becomes $\theta$, and the measure is $dx = 2a\tan\theta\sec^2\theta\,d\theta = d(a\tan^2\theta)$.

Step 2: Integrate:
So $I = \int\theta\,d(a\tan^2\theta) = a\theta\tan^2\theta - a\int\tan^2\theta\,d\theta$. Using $\tan^2\theta = \sec^2\theta - 1$ gives $\int\tan^2\theta\,d\theta = \tan\theta - \theta$.
\[ I = a\theta\tan^2\theta - a\tan\theta + a\theta + c \]
Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]