Find the value of $$\int_0^\pi\mathrm{e}^{\mathrm{e}^{\cos\left(x\right)}}\cos\left(\sin\left(x\right)\right)\cos\left(\mathrm{e}^x\sin\left(\sin\left(x\right)\right)\right)dx$$
How to solve this integral$?$ I tried using $u=\sin x$ substitution but it unnecessary made the problem more complicated, for instance $e^x$ will now be $e^{\arcsin u}.$ I also tried to calculate it with integral calculator but it failed. Any help is greatly appreciated.
Best Answer
Following up on @Random Variable's comment, spent quite some time trying to understand the actual question;
$$I=\int_0^\pi\mathrm{e}^{\mathrm{e}^{\cos\left(x\right)}}\cos\left(\sin\left(x\right)\right)\cos\left(\mathrm{e}^x\sin\left(\sin\left(x\right)\right)\right)dx$$
But it is not ending nicely, however if it were;
$$\int_0^\pi e^{e^{cosx}cos(sinx)}cos(e^{cosx}sin(sinx))\,dx$$
The integral can be evaluated and gives a nice answer. (I have a feeling that this was the actual question you intended to ask)
$$I=Re\left[\int_0^\pi e^{e^{cosx}cos(sinx)}e^{ie^{cosx}sin(sinx)}\,dx\right]$$
$$I=Re\left[\int_0^\pi e^{e^{cosx}(cos(sinx)+isin(sinx)}\,dx\right]$$
$$I=Re\left[\int_0^\pi e^{e^{cosx}ie^{sinx}}\,dx\right]$$
$$I=Re\left[\int_0^\pi e^{e^{cosx}+i{sinx}}\,dx\right]$$
$$I=Re\left[\int_0^\pi e^{e^{e^{ix}}}\,dx\right]$$
$$I=Re\left[\int_0^\pi \sum_{n=0}^{\infty} \frac{(e^{e^{ix}})^n}{n!}\,dx\right]$$
$$I=Re\left[\sum_{n=0}^{\infty}\frac{1}{n!} \int_0^{\pi}\sum_{k=0}^{\infty} \frac{(ne^{ix})^k}{k!}\,dx\right]$$
$$I=\sum_{n=0}^{\infty}\frac{1}{n!} \sum_{k=0}^{\infty} \frac{(n)^k}{k!}\int_0^{\pi} cos(kx) \,dx$$
$$I=\sum_{n=0}^{\infty}\frac{1}{n!} \sum_{k=0}^{\infty} \frac{(n)^k}{k!} \frac{sin{\pi k}}{k}$$
$$I=\sum_{n=0}^{\infty}\frac{1}{n!}\left[\pi+\sum_{k=1}^{\infty} \frac{(n)^k}{k!} \frac{sin{\pi k}}{k} \right]$$ $$I=\sum_{n=0}^{\infty}\frac{1}{n!}\left[\pi\right]=\pi e $$