Question

$\int e^{si{n}^{-1}x}[1+\frac{x}{\sqrt{1-x^2}}]$ dx $=$

• A. $x{e}^{si{n}^{-1}x}+c$
• B. $e^{si{n}^{-1}x}+c$
• C. $\frac{1}{\sqrt{1-x^2}}{e}^{{\sin }^{-1}x}+c$
• D. $x^2{e}^{si{n}^{1}X}+c$

Answer 1

The correct option is A $x{e}^{si{n}^{-1}x}+c$

$I=\int e^{si{n}^{-1}x}[1+\frac{x}{\sqrt{1-x^2}}]dx$
Put ${\sin }^{-1}x=t$
$\Rightarrow x=\sin t$
$\Rightarrow dx=\cos tdt$
So, $I=\int e^t(\sin t+\cos t)dt$
$\Rightarrow I=(\sin t+\cos t)e^t-\int (\cos t-\sin t)e^{t}dt$
$\Rightarrow I=(\sin t+\cos t)e^t-(\cos t-\sin t)e^t-\int e^t(\sin t+\cos t)dt$
$\Rightarrow 2I=2\sin t{e}^t+C$
$\Rightarrow I=x{e}^{{\sin }^{-1}x}+C$