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: Substitution:
Let \(x = a\tan^2\theta\), so \(\theta = \tan^{-1}\sqrt{\frac xa}\). Then \(\frac{a + x}{x} = \frac{\sec^2\theta}{\tan^2\theta} = \frac{1}{\sin^2\theta}\), so \(\sqrt{\frac{a+x}{x}} = \operatorname{cosec}\theta\). Hence \(\operatorname{cosec}^{-1}\sqrt{\frac{a+x}{x}} = \theta\).
Also \(dx = 2a\tan\theta\sec^2\theta\,d\theta\).

Step 2: Integrate by parts:
\[ I = \int\theta\cdot2a\tan\theta\sec^2\theta\,d\theta \]
Take \(u = \theta\) and \(dv = 2a\tan\theta\sec^2\theta\,d\theta\), so \(v = a\tan^2\theta\).
\[ I = a\theta\tan^2\theta - a\int\tan^2\theta\,d\theta = a\theta\tan^2\theta - a(\tan\theta - \theta) + c \]

Step 3: Simplify:
\[ I = a\theta\tan^2\theta - a\tan\theta + a\theta + c \]
This matches option (B), which has \(+a\theta\). Options with \(-a\theta\) come from a sign error in \(\int\tan^2\theta\,d\theta = \tan\theta - \theta\).

Final Answer:
The integral is \(a\theta\tan^2\theta - a\tan\theta + a\theta + c\), option (B). \[ \boxed{a\theta\tan^2\theta - a\tan\theta + a\theta + c} \]