Evaluate $\int_{}^{}e^{(\sin^{-1} x)}dx$

Question

Evaluate the integral :

$\int_{}^{}e^{(\sin^{-1} x)}dx$

I started solving this problem with Integration By Parts technique, setting $dv=dx$ and $u=e^{(\sin^{-1} x)}$

But I got an integral more complicated in the next step

Can this integral be solved by this technique?

Answer 1

let $x=sin(u)$

$dx=\cos(u) du$ by using by part to get $$\int e^u \cos u du=e^u\sin u+e^u\cos u-\int e^u\cos u du+C$$ $$2\int e^u \cos u du=e^u\sin u+e^u\cos u+C $$

$$\int e^u(\cos u)\ du=\frac{1}{2}e^{u} (\sin u+\cos u)+C_1$$ $$\int e^u(\cos u)\ du=\frac{1}{2}e^{u} (\sin u+\sqrt{1-\sin u^2})+C_1$$ $$=\frac{1}{2}e^{sin^{-1} x} (x+\sqrt{1-x^2})+C_1$$

Answer 2

Substitute $x= \sin\theta$. Can you solve it now?

Answer 3

Let $x=sin(\theta)$. Then $dx=\cos(\theta) d\theta$. Then the integral is $$I=\int e^{\theta}cos(\theta)d\theta$$ Let $u=e^{\theta}$, $dv=\cos(\theta)$, so $$I=e^{\theta}\sin(\theta)-\int e^{\theta}\sin(\theta)d\theta$$ Then $u=e^{\theta}$, $dv=\sin(\theta)$, $$I=e^{\theta}\sin(\theta)+e^{\theta}\cos(\theta)-I$$ So then $$I=\frac{1}{2}\left(e^{\theta}\sin(\theta)+e^{\theta}\cos(\theta)\right)+C$$ Simply plug back in the original substitution and use the Pythagorean identity for $\cos(\theta)$.