Find the $\space \displaystyle\int e^{\sin^{-1}x}~\mathrm dx$ .
I started by making a substitution. Let $u=\sin^{-1}x$, and so one can conclued that:
$\begin{align}1)&\mathrm du=\displaystyle\frac{1}{\sqrt{1-x^2}}\mathrm dx\\2)&x=\sin u
\end{align}$
So, the integral stays:
$\begin{align}\int e^{\sin^{-1}x}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \mathrm dx &=\int e^u\sqrt{1-\sin^2u}~\mathrm du=\int e^u\sqrt{\cos^2u}~\mathrm du=\int e^u\cos u~\mathrm du\end{align}$
Now, I tryed integration by parts but I could't managed. In Wolfram there is a complicated formula that I never heard about. Is there an intuitive way to finish this integral? Thanks.
Best Answer
Two consecutive integrations by parts should give you an equation satisfied by the primitive. More specifically, $$ \int_{-\infty}^a e^u\cos u\,\mathrm du = e^a\cos a + \int_{-\infty}^a e^u\sin u\,\mathrm du.$$
A second integration by parts gives an equation for $\int_{-\infty}^a e^u\cos u\,\mathrm du$.