Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left[\frac{\sqrt{1-x^2}{\sin }^{-1}x+1}{\sqrt{1-x^2}}\right]dx \\ =\int e^x\left[{\sin }^{-1}x+\frac{1}{\sqrt{1-x^2}}\right]dx \\ \mathrm{Here},f\left(x\right)={\sin }^{-1}x \\ \Rightarrow f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}} \\ \mathrm{Put}{e}^{x}f\left(x\right)=t \\ \Rightarrow e^x{\sin }^{-1}x=t \\ Diffbothsidesw.r.tx \\ \left(e^x{\sin }^{-1}x+e^x\times \frac{1}{\sqrt{1-x^2}}\right)dx=dt \\ \therefore I=\int dt \\ =t+C \\ =e^x{\sin }^{-1}x+C \end{gathered}$$