Integral $\int-\sin x\mathrm e^{-\sin x}$ I want to integrate the function $$-\sin x\mathrm{e}^{-\sin x}.$$
I tried integration by parts with $u=-\sin x$ but I got another difficult integration which is $$\int (\cos x)^{2}\mathrm{e}^{-\sin x} \,\mathrm{d}x.$$
Could someone help me out? 
 A: I do not think that you could find the antiderivative.
Set $\sin(x)=t$ to make the integrand $-t\, e^{-t}$ and then
$$-t e^{-t}=-\sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n!}$$ making
$$\int-\sin (x)\,\mathrm e^{-\sin (x)}\,dx=\sum_{n=0}^\infty  \frac{(-1)^{n+1}}{n!}\int \sin^{n+1}(x)\,dx$$ 
Edit
Since, in comments, the problem is said for the integral between $0$ and $2\pi$, let us use
$$\int_0^{2\pi} \sin^{n+1}(x)\,dx=-\sqrt{\pi }\frac{ \left((-1)^n-1\right) \Gamma \left(\frac{n+2}{2}\right)}{\Gamma
   \left(\frac{n+3}{2}\right)}$$ which are equal to $0$ if $n$ is even.
This makes by the end$$-\int_0^{2\pi}\sin (x)\,\mathrm e^{-\sin (x)}\,dx=\sum_{n=0}^\infty \frac{  2^{-2 n}\, \pi}{\Gamma (n+1) \Gamma (n+2)} =2 \pi  I_1(1)$$ where appears the modified Bessel function of the first kind.
Looking for explicit results, using the same approach as above for $\int_0^{\frac{n\pi}2} \sin^{n+1}(x)\,dx$, I only found some for
$$I_n=-\int_0^{\frac{n\pi}2} \sin(x)\,\mathrm e^{-\sin (x)}\,dx$$ the key ones being
$$\left(
\begin{array}{cc}
n & I_n \\
 1 & \frac{1}{2} \pi  (I_1(1)-\pmb{L}_{-1}(1)) \\
 2 & \pi  (I_1(1)-\pmb{L}_{-1}(1)) \\
 3 & \frac{1}{2} \pi  (3 I_1(1)-\pmb{L}_{-1}(1)) \\
 4 & 2 \pi  I_1(1)
\end{array}
\right)$$ where also appears the modified Struve function.
A: $$\int\cos x\,e^{\sin x}dx$$ is elementary. Using the chain rule, it is easy to see that the antiderivative is $$e^{\sin x}.$$
On the opposite,
$$\int\sin x\,e^{\sin x}dx$$ has no closed-form expression.
A: $$
\begin{align}
-\int_0^{2\pi}\sin(x)\,e^{-\sin(x)}\,\mathrm{d}x
&=\int_0^{2\pi}\sin(x)\,e^{\sin(x)}\,\mathrm{d}x\\
&=\int_0^{2\pi}\sin(x)\,\sinh(\sin(x))\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\int_0^{2\pi}\frac{\sin^{2k+2}(x)}{(2k+1)!}\,\mathrm{d}x\\
&=2\pi\sum_{k=0}^\infty\frac1{4^{k+1}}\binom{2k+2}{k+1}\frac1{(2k+1)!}\\
&=\pi\sum_{k=0}^\infty\frac1{4^kk!(k+1)!}\\[6pt]
&=2\pi\operatorname{I}_1(1)\\[15pt]
&=3.5509993784243618938
\end{align}
$$
$\operatorname{I}_1$ is a modified Bessel function of the first kind. In any case, the sum converges quite quickly.
A: $-\int\mathrm e^{-\sin x}\sin x~dx=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$
For $n$ is any non-negative integer,
$\int\sin^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$
This result can be done by successive integration by parts.
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n+2)!x}{4^{n+1}(2n+1)!((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}(2n+1)!((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{2(n+1)x}{4^{n+1}((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(n+1)(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x}{2^{2n+1}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\sin^{2k+1}x\cos x}{2^{2n-2k+1}n!(n+1)!(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
