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Evaluation of \(\displaystyle \int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx\)

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

Proof:
We are making use of a theorem stating that:
Theorem: Let \( f \) be a periodic function with a period \( T \neq 0 \). Then:
$$ \int_{a+mT}^{b+nT}f(x)\,dx=\int_{a}^{b}f(x)\,dx+\left ( n-m \right )\int_{0}^{T}f(x)\,dx$$
where \( m, n \in \mathbb{Z} \).

We also see that the integrand function is periodic with a period \(T=2\pi \neq 0 \), since \( \displaystyle \frac{1}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}=\frac{1}{\left ( 1+2^{\sin \left ( x+2\pi \right )} \right )\left ( 1+2^{\cos \left ( x+2\pi \right )}\right )} \). If we denote the integrand function as \(f(x) \) and at the same time make use of the theorem our integral yields: $$\int_{\pi/4+3\cdot 2\pi}^{5\pi/4+6\cdot 2\pi}f(x)\,dx=\int_{\pi/4}^{5\pi/4}f(x)\,dx+\left ( 6-3 \right )\int_{0}^{2\pi}f(x)\,dx= \int_{\pi/4}^{5\pi/4}f(x)\,dx+3\int_{0}^{2\pi}f(x)\,dx$$
Let us denote each integral \(A \) and \( B \) respectively, thus \( \displaystyle A=\int_{\pi/4}^{5\pi/4}f(x)\,dx, \quad B=3\int_{0}^{2\pi}f(x)\,dx \).

We'll begin from integral \(B \). We tear the integral apart into \(2 \) integrals, thus: $$B=3\left ( \int_{0}^{\pi}f(x) +\int_{\pi}^{2\pi}f(x)\right )\,dx$$ Now $$\begin{aligned}
\int_{\pi}^{2\pi}f(x)\,dx &=\int_{\pi}^{2\pi}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )} \\
&\overset{u=x-\pi}{=\! =\! =\! =\!} \int_{0}^{\pi}\frac{2^{\sin u}2^{\cos u}}{\left ( 1+2^{\sin u} \right )\left ( 1+2^{\cos u} \right )}\,du\\
\end{aligned}$$ Thus $$B=3\left ( \int_{0}^{\pi}f(x)+\int_{\pi}^{2\pi}f(x) \right )\,dx=3\int_{0}^{\pi}\frac{1+2^{\sin x}2^{\cos x}}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}\,dx$$ Apllying the substitution \( u=\pi-x \) we finally get that \( B \) equals \( \displaystyle \frac{3\pi}{2} \).

Now back to the \(A \) 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 apply the substitution \( u= x-\dfrac{\pi}{2} \) so the integral yields \( \displaystyle \int_{\pi/4}^{3\pi/4}\frac{2^{\sin x}}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}dx \). Add it to the first one so that \(A \) eventually can be written as: $$B=\int_{\pi/4}^{3\pi/4}\frac{1+2^{\sin x}}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}\,dx =\int_{\pi/4}^{3\pi/4}\frac{dx}{1+2^{\cos x}}=\frac{\pi}{2}\cdot \frac{1}{1+2^{\cos \frac{\pi}{2}}}=\frac{\pi}{4}$$ Now we can see that we have evaluated the given integral. Add \(A \) and \( B\) to get the final result:
$$\boxed{\displaystyle \int_{25\pi/4}^{53\pi/4}\frac{dx}{\left ( 1+2^{\sin x} \right )\left ( 1+2^{\cos x} \right )}dx=\frac{7\pi}{4}}$$

T:

Fascinating but extremely difficult and tedious....