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

The correct answer is 26.

$I=\int_{0}^{\pi}\frac{5^{cosx}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)}{1+5^{cosx}}dx$

$I=\int_{0}^{\pi}\frac{5^{-cosx}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)}{1+5^{-cosx}}dx$

$2I=\int_{0}^{\pi}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)dx$

$\not{2}I=\not{2}\int_{0}^{\frac{\pi}{2}}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)dx$

$I=\int_{0}^{\frac{\pi}{2}}(1+sinx(-sin3x)+sin^{2}x-sin^{3}xsin3x)dx$

$2I=\int_{0}^{\frac{\pi}{2}}(3+cos4x+cos^{3}xcos3x-sin^{3}xsin3x)dx$

$2I=\int_{0}^{\frac{\pi}{2}}3+cos4x+(\frac{cos3x+3cosx}{4})cos3x-sin3x(\frac{3sinx-sin3x}{4})dx$

$2I=\int_{0}^{\frac{\pi}{2}}(3+cos4x+(\frac{1}{4})+\frac{3}{4}cos4x)dx$

$2I=\frac{13}{4}\times\frac{\pi }{2}+\frac{7}{4}(\frac{sin4x}{4})_{0}^{\frac{\pi }{2}}\Rightarrow I=\frac{13\pi }{16}$