If ∫0π 5cos⁡x1+cos⁡xcos⁡3x+cos2⁡x+cos3⁡xcos⁡3xdx1+5cos⁡x=kπ16, then k is equal to ____

I=∫0π 5cos⁡x1+cos⁡xcos⁡3x+cos2⁡x+cos3⁡xcos⁡3x1+5cos⁡xdx……(1)=∫0π 5−cos⁡x1+cos⁡xcos⁡3x+cos2⁡x+cos3⁡xcos⁡3x1+5−cos⁡xdx……(2)(1)+(2)⇒2I=∫0π 5cos⁡x+15cos⁡x+11+cos⁡xcos⁡3x+cos2⁡x+cos3⁡xcos⁡3xdx⇒2I=∫0π 1+12(cos⁡4x+cos⁡2x)+12(1+cos⁡2x)dx+∫0π cos3⁡x4cos3⁡x−3cos⁡xdx=∫0π 1+12cos⁡4x+12cos⁡2x+12+cos⁡2x2dx+4∫0π cos6⁡xdx−3∫0π cos4⁡xdx=x+sin⁡4x8+sin⁡2x4+x2+sin⁡2x40π+856×34×12×π2−634×12×π2=π+0+0+π2+0+5π4−9π8=3π2+π8=12π+π8=13π82I=13π8⇒I=13π16∴k=13

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If $\int_{0}^{π} \frac{5^{\cos x} (1 + \cos xcos 3 x + \cos^{2} x + \cos^{3} xcos 3 x) dx}{1 + 5^{\cos x}} = \frac{kπ}{16}$ , then $k$ is equal to ____

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answer is 13.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{π} \frac{5^{\cos x} (1 + \cos xcos 3 x + \cos^{2} x + \cos^{3} xcos 3 x)}{1 + 5^{\cos x}} dx \dots \dots (1) \\ = \int_{0}^{π} \frac{5^{- \cos x} (1 + \cos xcos 3 x + \cos^{2} x + \cos^{3} xcos 3 x)}{1 + 5^{- \cos x}} dx \dots \dots (2) \\ (1) + (2) \\ \Rightarrow 2 I = \int_{0}^{π} \frac{(5^{\cos x} + 1)}{5^{\cos x} + 1} (1 + \cos xcos 3 x + \cos^{2} x + \cos^{3} xcos 3 x) dx \\ \Rightarrow 2 I = \int_{0}^{π} [1 + \frac{1}{2} (\cos 4 x + \cos 2 x) + \frac{1}{2} (1 + \cos 2 x)] dx + \int_{0}^{π} \cos^{3} x (4 \cos^{3} x - 3 \cos x) dx \\ = \int_{0}^{π} [1 + \frac{1}{2} \cos 4 x + \frac{1}{2} \cos 2 x + \frac{1}{2} + \frac{\cos 2 x}{2}] dx + 4 \int_{0}^{π} \cos^{6} xdx - 3 \int_{0}^{π} \cos^{4} xdx \\ = {(x + \frac{\sin 4 x}{8} + \frac{\sin 2 x}{4} + \frac{x}{2} + \frac{\sin 2 x}{4})}_{0}^{π} + 8 (\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{π}{2}) - 6 (\frac{3}{4} \times \frac{1}{2} \times \frac{π}{2}) \\ = π + 0 + 0 + \frac{π}{2} + 0 + \frac{5 π}{4} - \frac{9 π}{8} \\ = \frac{3 π}{2} + \frac{π}{8} = \frac{12 π + π}{8} = \frac{13 π}{8} \\ 2 I = \frac{13 π}{8} \Rightarrow I = \frac{13 π}{16} \\ ∴ k = 13 \end{array}$