Evaluating integral with $e^{\sin x}$ I had this integral $ \int e^{\sin(x)} {\sin(2x)} dx$ 
I tried to split it up using integration by parts but I can't evaluate integral of $e^{\sin x}$
 A: Write $\sin 2x = 2 \sin x \cos x$. The substitution $u = \sin x$ will yield $$2 \int e^u u \, du$$ which you can integrate by parts.
A: By parts works perfectly,
$$\int2\sin(x)\left(\cos(x)e^{\sin(x)}\right)dx=2\sin(x)e^{\sin(x)}-2\int\cos(x)e^{\sin(x)}dx\\
=2\sin(x)e^{\sin(x)}-2e^{\sin(x)}$$
A: Note that $\sin(2x) = 2\sin(x)\cos(x)$. Then make the $u$-sub, $u = \sin(x)$. The integral becomes
$$ 2\int u e^u du.$$
Now integrate by parts.
A: Let $u = \sin x$ and $dv = e^x dx$, then $du = \cos x$ and $v = e^x$, so 
$$\int \left(e^x \sin x\right)dx = e^x \sin x - \int \left(e^x \cos x\right) dx $$
Now integrate by parts again: 
This time with $u = \cos x$ and $dv = e^x dx$, $du = -\sin x$ dx, $v = e^x$ 
So $$\int \left(e^x \sin x\right) dx = e^x \sin(x) - e^x \cos(x) - \int \left(e^x sin x\right) dx $$
We can add $\int \left(e^x \sin x\right) dx$ to both sides to get $$2 \int\left(e^x \sin x\right) dx = e^x \left(\sin x - \cos x\right)$$ and 
$$\int \left(e^x \sin x\right) dx = \frac{e^x(\sin x - \cos x)}{2}$$
