Question

Find the integer closest to ${\int }_0^{2\pi }\frac{\pi dx}{(1+2^{sinx})(1+2^{cosx})}$.

Answer 1

$f\left(x\right)={\int }_0^{2\pi }\frac{\pi dx}{(1+2^{\sin x})(1+2^{\cos x})}=I$
$I=f(2\pi -x)$
$={\int }_0^{2\pi }\frac{\pi dx}{(1+2^{\sin (2\pi -x)})(1+2^{\cos (2\pi -x)})}$
$2I={\int }_0^{2\pi }\frac{\pi dx}{(1+2^{\sin x})(1+2^{\cos x})}+{\int }_0^{2\pi }\frac{\pi {\cdot 2}^{\sin x}\cdot dx}{(1+2^{\sin x})(1+2^{\cos x})}$
$2I=\pi {\int }_0^{2\pi }\frac{(1+2^{\sin x})dx}{(1+2^{\cos x})(1+2^{\sin x})}=\pi {\int }_0^{2\pi }\frac{dx}{(1+2^{\cos x})}$
$-1\le \cos x\le 1$
$\frac{1}{2}\le 2^{\cos x}\le 2^1$
$\frac{3}{2}\le 2^{\cos x}+1\le 3$
$\frac{1}{3}\le \frac{1}{2^{\cos x}+1}\le \frac{2}{3}$
${\int }_0^{2\pi }\frac{\pi }{3}dx\le {\int }_0^{2\pi }\frac{\pi dx}{2^{\cos x}+1}\le {\int }_0^{2\pi }\frac{2\pi }{3}dx$
$\frac{2{\pi }^2}{3}\le 2I\le \frac{4{\pi }^2}{3}$
$\frac{{\pi }^2}{3}\le I\le \frac{2{\pi }^2}{3}$
$I$ can take integer values of 3,4,5,6
$I$ can take '6' integer closest to it