Prove the following equality: $\int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi$

Question

Prove the following equality: $$ \int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi $$

Answer 1

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.

Answer 2

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}):

$$\begin{align} e^{4\cos(t)}\cos(4\sin(t))&=e^{4\cos(t)}\frac12(e^{i4\sin(t)}+e^{-i4\sin(t)})\\ &= \frac12(e^{4(\cos(t)+i\sin(t))}+e^{4(\cos(t)-i\sin(t))})\\ &=\frac12(e^{4e^{it}}+e^{4e^{-it}})\end{align}$$

Answer 3

$e^{4\cos t}\cos(4\sin t)$ 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.