Find $\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$ Find the value of: $$I=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$$
First, I rewrote the limits as the function goes from $\frac{\pi}{4}$ to $\frac{9\pi}{4}$. Now the integral with the reduced limits has the value $\frac{I}{8}$. I used standard properties of definite integrals like converting $f(x)$ to $f(a+b-x)$ where $a,b$ are the lower and upper limits of the integral. Considering symmetry of the trigonometric functions, I believe I can even change the limits to $0$ to $2\pi$. However, this does not simplify the problem as much as I would like. Please advice.   
 A: \begin{align*}
I&=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \\
&= 8\int_{-\pi}^{\pi}  \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \tag{1}\\
&= 8\int_{-\pi}^{\pi}\frac{dx}{\left(1+2^{\cos (-x)}\right)\left(1+2^{\sin (-x)}\right)} \tag{2}\\
&=8\int_{-\pi}^{\pi} \frac{dx}{\left(1+2^{\cos x}\right)\left(1+2^{-\sin x}\right)} \tag{3}\\
&= 8\int_{-\pi}^{\pi} \frac{2^{\sin x} dx}{\left(1+2^{\cos x}\right)\left(1+2^{\sin x}\right)} \tag{4}\\
\implies 2I &= 8\int_{-\pi}^{\pi} \frac{1+2^{\sin x} dx}{\left(1+2^{\cos x}\right)\left(1+2^{\sin x}\right)} \tag{5}\\
&= 8 \int_{-\pi}^{\pi} \frac{dx}{1+2^{\cos x}} \\
\implies I &= 4 \int_{-\pi}^{\pi} \frac{dx}{1+2^{\cos x}}  \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{\cos x}} \tag{6} \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{\cos (\pi-x)}} \tag{7} \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{-\cos x}} \tag{8} \\
&= 8 \int_{0}^{\pi} \frac{2^{\cos x} dx}{1+2^{\cos x}}  \tag{9} \\
\implies 2I &= 8\int_{0}^{\pi} \frac{1+2^{\cos x} dx}{1+2^{\cos x}} \tag{10} \\
\implies I &= 4 \int_{0}^{\pi} dx \\
&= \color{red}{4 \pi}
\end{align*}
In $(1)$, we have used the periodicity of $\sin x$ and $\cos x$ and brought the  limits  down to more familiar territory.
In $(2)$, we have used the $a+b-x$ trick.
In $(3)$ we used that $\sin x$ is odd and $\cos x$ is even.
In $(5)$ we have added $(1)$ to $(4)$
In $(6)$ we use that $\cos x$ is even.
In $(7)$, $a+b-x$ again.
In $(8)$, $\cos(\pi-x) = -\cos x$
In $(10)$, we do $(6) + (9)$
