Question

If $\int \limits_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x}{1+5^{\cos x}}=\frac{ k \pi}{16}$, then $k$ is equal to

Answer 1

$ I=\int\limits_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{\cos x}} d x$
$ I=\int\limits_0^\pi \frac{5^{-\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{-\cos x}} d x $
$ 2 I=\int\limits_0^\pi\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x$
$\not{2} I=\not 2 \int\limits_0^{\frac{\pi}{2}}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x$
$I =\int\limits_0^{\frac{\pi}{2}}\left(1+\sin x (-\sin 3 x )+\sin ^2 x -\sin ^3 x \sin 3 x \right) dx$
$ 2 I =\int\limits_0^{\frac{\pi}{2}}\left(3+\cos 4 x +\cos ^3 x \cos 3 x -\sin ^3 x \sin 3 x \right) dx $
$ 2 I =\int\limits_0^{\frac{\pi}{2}} 3+\cos 4 x +\left(\frac{\cos 3 x+3 \cos x }{4}\right) \cos 3 x -\sin 3 x \left(\frac{3 \sin x-\sin 3 x}{4}\right) d x$
$2 I=\int\limits_0^{\frac{\pi}{2}}\left(3+\cos 4 x+\frac{1}{4}+\frac{3}{4} \cos 4 x\right) d x $
$ 2 I=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4 x}{4}\right)_0^{\frac{\pi}{2}} \Rightarrow I =\frac{13 \pi}{16}$