Question

${\int }_0^{1/2}{e}^x[si{n}^{-1}x+\frac{1}{\sqrt{1-x^2}}]$ dx $=$

• A. $\frac{e^4}{4}$
• B. $\frac{\pi \sqrt{e}}{6}$
• C. $\frac{\sqrt{\pi e}}{4}$
• D. $\frac{\pi \sqrt{e}}{2}$

Answer 1

The correct option is B $\frac{\pi \sqrt{e}}{6}$

${\int }_0^{\frac{1}{2}}{e}^x[{\sin }^{-1}x+\frac{1}{\sqrt{1-x^2}}]dx$

Consider, $\int e^x({\sin }^{-1}x+\frac{1}{\sqrt{1-x^2}})dx$

It is in the form of $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$

$=e^{x}f\left(x\right)+c$

$\therefore {\int }_0^{\frac{1}{2}}{e}^x({\sin }^{-1}x+\frac{1}{\sqrt{1-x^2}})dx={[e^x{\sin }^{-1}x+c]}_0^{\frac{1}{2}}$

$=e^{\frac{1}{2}}{\sin }^{-1}\left(\frac{1}{2}\right)-\left(e^0{\sin }^{-1}0\right)$

$=e^{\frac{1}{2}}\cdot \frac{\pi }{6}=\frac{\pi \sqrt{e}}{6}$