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