Question

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

Answer 1

$$\text{ Let I }=\int e^x\left[\frac{\sqrt{1-x^2}{\sin }^{-1}x+1}{\sqrt{1-x^2}}\right]dx$$

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

$$\text{ Here},f(x)={\sin }^{-1}x$$

$$\Rightarrow f^{\prime }(x)=\frac{1}{\sqrt{1-x^2}}$$

$${\text{ Put e}}^{x}f(x)=t$$

$$\Rightarrow e^x{\sin }^{-1}x=t$$

$$\text{ Diff both sides w . r . t x}$$

$$(e^x{\sin }^{-1}x+e^x\times \frac{1}{\sqrt{1-x^2}})dx=dt$$

$$\therefore I=\int dt$$

$$=t+C$$

$$=e^x{\sin }^{-1}x+C$$