Question

If $ f(x)=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} $ and $ g(x)={{e}^{{{\sin }^{-1}}x}}, $ then $ \int{f(x)g(x)}dx $ is equal to

• A. $ {{e}^{{{\sin }^{-1}}x}}({{\sin }^{-1}}x-1)+c $
• B. $ {{e}^{{{\sin }^{-1}}x}}+c $
• C. $ {{e}^{{{({{\sin }^{-1}}x)}^{2}}}}+c $
• D. $ {{e}^{2{{\sin }^{-1}}x}}+c $
• E. $ {{e}^{{{\sin }^{-1}}x}}{{\sin }^{-1}}x+c $

Answer 1

Answer: (A)

Given that, $ f(x)=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} $ and $ g(x)={{e}^{{{\sin }^{-1}}x}} $
$ \therefore $ $ \int{f(x)}g(x)dx=\int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{{e}^{{{\sin }^{-1}}x}}dx $
Let $ {{\sin }^{-1}}x=t $ and, $ \frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt $
$ \Rightarrow $ $ \int{f(x)g(x)}dx=\int{t{{e}^{t}}}dt=t{{e}^{t}}-{{e}^{t}}+c $
$={{e}^{{{\sin }^{-1}}x}}({{\sin }^{-1}}x-1)+c $