Evaluation of $-\int e^{\cos(t)}\sin(\sin(t)+t)\,dt $ How would I integrate this:
$$-\int e^{\cos(t)}\sin(\sin(t)+t)\,dt $$
I have tried several methods but can't seem to work this out.
 A: If you are allowed to use complex number, this integral can be integrated by
repeat application of the Euler's formula
$$e^{i\theta} = \cos(\theta) + i\sin\theta$$
Up to integration constant, the integral is equal to:
$$\begin{align}\int -e^{\cos t}\sin(\sin t + t) dt
&= - \int e^{\cos t}\Im\left[e^{i(\sin t + t)}\right] dt
\stackrel{\color{blue}{[1]}}{=} - \Im \left[\int e^{\cos t + i(\sin t + t)} dt \right]\\
&=  - \Im \left[\int e^{e^{it}} e^{it} dt\right]
=  \Im \left[ i\int e^{e^{it}} d e^{it}\right]\\
&= \Re\left[ e^{e^{it}} \right]
= \Re\left[ e^{\cos t + i\sin t} \right]\\
&= \, e^{\cos t} \cos(\sin t)
\end{align}
$$
Notes


*

*$\color{blue}{[1]}$ - At this step, the integral is an ordinary integral of a complex valued function along the real axis. There is no issue of moving the $\Im[\cdots]$ outside the integral sign.

A: Note that $$e^{\cos t} \sin(\sin(t) + t) = e^{\cos t} \left( \sin(\sin t)\cos t + \cos(\sin t)\sin t \right)$$ which is equal to $$-\left(e^{\cos t}(\cos t)'\cos(\sin t) + e^{\cos t}(\cos(\sin t))'\right)$$
Got it from here?
