A tough integral

Evaluate the integral:
$$\mathcal{J}=\int_{25\pi/4}^{53\pi/4}\frac{{\rm d}x}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}$$

Answer: $\displaystyle \int_{25\pi/4}^{53\pi/4}\frac{{\rm d}x}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}=\frac{7\pi}{4}$


Proof:

 
We are making use of a theorem stating that if a function $f$ is periodic with a period $T \neq 0$ then:

$$\int_{a+mT}^{b+nT}f(x)\,{\rm d}x=\int_{a}^{b}f(x)\,{\rm d}x+(n-m)\int_{0}^{T}f(x)\,{\rm d}x$$

whereas $m,n \in \mathbb{Z}$.

We now note that the integrand is periodic with a period of $2\pi$ since:
$$\frac{1}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}=\frac{1}{\left ( 1+2^{\sin \left ( x+2\pi \right )} \right )\left ( 1+2^{\cos \left ( x+2\pi \right )}\right )}$$

Hence, denoting the integrand as $f(x)$ then from the thoerem yields that:

$$\int_{\pi/4+3\cdot 2\pi}^{5\pi/4+6\cdot 2\pi}f(x)\,{\rm d}x=\int_{\pi/4}^{5\pi/4}f(x)\,{\rm d}x+\left ( 6-3 \right )\int_{0}^{2\pi}f(x)\,{\rm d}x= \int_{\pi/4}^{5\pi/4}f(x)\,{\rm d}x+3\int_{0}^{2\pi}f(x)\,{\rm d}x$$

Let us denote each integral as $\mathcal{A}$ and $\mathcal{B}$ respectively. We begin by the second integral where we tear it apart into two integrals, hence it is written as:

$$\mathcal{B}=3\left ( \int_{0}^{\pi}f(x)\,{\rm d}x +\int_{\pi}^{2\pi}f(x)\,{\rm d}x\right )$$

By applying the substitution $u=x-\pi$ at the second integral we have that:

$$\begin{align*} \int_{\pi}^{2\pi}f(x)\,dx &=\int_{\pi}^{2\pi}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )} \\ &\overset{u=x-\pi}{=\! =\! =\! =\!} \int_{0}^{\pi}\frac{2^{\sin u}2^{\cos u}}{\left ( 1+2^{\sin u} \right )\left ( 1+2^{\cos u} \right )}\,du\\ \end{align*}$$

and by adding to the first integral we get that:

$$\mathcal{B}=3\left ( \int_{0}^{\pi}f(x)\,{\rm d}x+\int_{\pi}^{2\pi}f(x)\,{\rm d}x \right )=3\int_{0}^{\pi}\frac{1+2^{\sin x}2^{\cos x}}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}\,{\rm d}x$$

Finally by applying the substitution $u=\pi-x$ we get that $\mathcal{B}$ equals $\displaystyle \frac{3\pi}{2}$

Back to the $\mathcal{A}$ integral. Successively we have:

$$\begin{aligned} A=\int_{\pi/4}^{5\pi/4}f(x)\,dx &=\int_{\pi/4}^{3\pi/4}f(x)\,dx+\int_{3\pi/4}^{5\pi/4}f(x)\,dx \\ &= \int_{\pi/4}^{3\pi/4}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}+\int_{3\pi/4}^{5\pi/4}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )} \\ \end{aligned}$$

At the second integral we apply the sub $x =\pi/2 -u$ and to what we find we add to the first integral , so that $\mathcal{A}$ can finally be written as:

$$\mathcal{A}=\int_{\pi/4}^{3\pi/4}\frac{1+2^{\sin x}}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}\,dx =\int_{\pi/4}^{3\pi/4}\frac{dx}{1+2^{\cos x}}=\frac{\pi}{2}\cdot \frac{1}{1+2^{\cos \frac{\pi}{2}}}=\frac{\pi}{4}$$

Adding both of them we , finally get the final result that the initial integral is:

$$\boxed{\displaystyle \int_{25\pi/4}^{53\pi/4}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}dx=\frac{7\pi}{4}}$$

which gives the desired result and with that the exercise comes to an end.