If ∫ 0 π 5 cos x 1 + cos x cos 3 x + cos 2 x + cos 3 x cos 3 x d x 1 + 5 cos x = k π 16 ,…

Given,

$I={\int }_0^{\pi }\frac{5^{\cos x}\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)}{1+5^{\cos x}}dx ....\left(1\right)$

Now using the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get,

$\Rightarrow I={\int }_0^{\pi }\frac{5^{-\cos x}\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)}{1+5^{-\cos x}}dx ....\left(2\right)$

Now adding above equations we get,

$2I={\int }_0^{\pi }\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)dx$

Now using the property ${\int }_0^{2a}f\left(x\right)dx=2{\int }_a^{a}f\left(x\right)dx \left\{\text{when }f\left(x\right)\text{ is even}\right\}$

$\Rightarrow 2I=2{\int }_0^{\frac{\pi }{2}}\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)dx .....\left(3\right)$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\left(1+\sin x\left(-\sin 3x\right)+{\sin }^{2}x-{\sin }^{3}x\sin 3x\right)dx ....\left(4\right)$

Now adding equation $\left(3\right) \& \left(4\right)$ we get,

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\left(3+\cos 4x+{\cos }^{3}x\cos 3x-{\sin }^{3}x\sin 3x\right)dx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}3+\cos 4x+\left(\frac{\cos 3x+3\cos x}{4}\right)\cos 3x-\sin 3x\left(\frac{3\sin x-\sin 3x}{4}\right)dx$

$\Rightarrow 2\mathrm{I}={\int }_0^{\frac{\pi }{2}}\left(3+\cos 4\mathrm{x}+\frac{1}{4}+\frac{3}{4}\cos 4\mathrm{x}\right)\mathrm{dx}$

$\Rightarrow 2\mathrm{I}=\frac{13}{4}\times \frac{\pi }{2}+\frac{7}{4}{\left(\frac{\sin 4\mathrm{x}}{4}\right)}_0^{\frac{\pi }{2}}$

$\Rightarrow \mathrm{I}=\frac{13\pi }{16}$,

Now on comparing with $\frac{k\mathrm{\pi }}{16}$, we get $k=13$