Prove the following equality: $\int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi$ Prove the following equality: 
$$
\int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi
$$
 A: It is easy to see that the integral equals 
$$\tag{$*$}\label{eq1}I=\frac12\int_0^\pi (e^{4e^{it}}+e^{4e^{-it}})\,dt.$$
The substitution $u=\pi-t$ yields
$$I=\frac12\int_0^\pi (e^{-4e^{iu}}+e^{-4e^{-iu}})\,du(\equiv\frac12\int_0^\pi (e^{-4e^{it}}+e^{-4e^{-it}})\,dt)$$
so
$$\begin{align}
2I&=I+I\\
&=\frac12\int_0^\pi (e^{4e^{it}}+e^{4e^{-it}})\,dt+\frac12\int_0^\pi (e^{-4e^{it}}+e^{-4e^{-it}})\,dt\\
&=\frac12\int_0^\pi (e^{4e^{it}}+e^{4e^{-it}}+e^{-4e^{it}}+e^{-4e^{-it}})\,dt\\
&=\frac12\int_0^\pi (e^{4e^{it}}+e^{-4e^{it}}+e^{4e^{-it}}+e^{-4e^{-iu}})\,dt\\
&=\int_0^\pi \left(\frac12(e^{4e^{it}}+e^{-4e^{it}})+\frac12(e^{4e^{-it}}+e^{-4e^{-it}})\right)\,dt\\
&=\int_0^\pi(\cosh(4e^{it})+\cosh(-4e^{it}))\,dt=2\int_0^\pi \cosh(4e^{it})\,dt
\end{align}$$
or 
$$I=\int_0^\pi\cosh(4e^{it})\,dt.$$
Now let us define the function
$$\phi(x):=\int_0^\pi \cosh(xe^{it})\,dt.$$
Differentiation with respect to $x$ yields
$$\phi'(x)=\int_0^\pi e^{it}\sinh(xe^{it})\,dt
=-\left.\frac{i \cosh \left(e^{i t} x\right)}{x}\right]_0^\pi=0$$
so $\phi(x)$ is constant. This implies that 
$$I=\phi(4)=\phi(0)=\int_0^\pi\cosh(0)\,dt=\pi.$$
EDIT (how to derive equation\eqref{eq1}): 
$$\begin{align}
e^{4\cos(t)}\cos(4\sin(t))&=e^{4\cos(t)}\frac12(e^{i4\sin(t)}+e^{-i4\sin(t)})\\
&=
\frac12(e^{4(\cos(t)+i\sin(t))}+e^{4(\cos(t)-i\sin(t))})\\
&=\frac12(e^{4e^{it}}+e^{4e^{-it}})\end{align}$$
A: $e^{4\cos t}\cos(4\sin t)$ is the real part of $e^{4\cos t}\cdot e^{4i\sin t}$, aka $\exp\left(4e^{it}\right)$. Since
$$ \exp\left(4 e^{it}\right) = \sum_{n\geq 0}\frac{4^n e^{nit}}{n!} \tag{1}$$
and $\int_{0}^{\pi}e^{nit}\,dt$ equals either $0$ or a purely imaginary number for any $n\geq 1$,
$$\int_{0}^{\pi}e^{4\cos t}\cos(4\sin t)\,dt = \sum_{n\geq 0}\frac{4^n}{n!}\text{Re}\int_{0}^{\pi}e^{nit}\,dt=\int_{0}^{\pi}1\,dx = \color{red}{\pi},\tag{2}$$
i.e. only the contribute given by $n=0$ really matters.
A: Hint :
Consider the other integral
$S=\int_0^\pi e^{4cos(t)}sin(4sin(t))dt$
and the sum
$C+iS=\int_0^\pi e^{4(cos(t)+isin(t))}dt$
$=\int_0^\pi e^{4e^{it}}dt$
put $4e^{it}=u$ or $t=-iln(u)+i2ln(2)$
the integral becomes
$i\int_{-1}^{1}\frac{e^udu}{u}$
I'll surely be downvoted.
