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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.

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1 Answer 1

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\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 $(6) + (9)$

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    $\begingroup$ Nicely done! +1 $\endgroup$
    – Mark Viola
    May 14, 2016 at 18:59

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