JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Definite Integrals

Writing $I$ for the given integral and replacing $x$ by $\pi -x$, the factor $5^{\cos x}$ becomes $5^{-\cos x}$ while the bracket $1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x$ stays unchanged, because $\cos x$ and $\cos 3x$ change sign together. Adding the original and transformed integrals, the exponential weights sum to $1$, so $2I={\int }_0^{\pi }\left(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x\right)dx$. Over $[0,\pi ]$, ${\int }_0^{\pi }1dx=\pi$, ${\int }_0^{\pi }\cos x\cos 3xdx=0$, ${\int }_0^{\pi }{\cos }^{2}xdx=\frac{\pi }{2}$, and with ${\cos }^{3}x=\frac{1}{4}(3\cos x+\cos 3x)$ the last term gives ${\int }_0^{\pi }{\cos }^{3}x\cos 3xdx=\frac{\pi }{8}$. Hence $2I=\pi +\frac{\pi }{2}+\frac{\pi }{8}=\frac{13\pi }{8}$, so $I=\frac{13\pi }{16}$, and matching with $\frac{k\pi }{16}$ gives $k=13$.