JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 18)
If $$\int\limits_0^\pi {{{{5^{\cos x}}(1 + \cos x\cos 3x + {{\cos }^2}x + {{\cos }^3}x\cos 3x)dx} \over {1 + {5^{\cos x}}}} = {{k\pi } \over {16}}} $$, then k is equal to _____________.
Answer
13
Explanation
$$
\begin{aligned}
& \mathrm{I}=\int_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{\cos x}} d x \\\\
& I=\int_0^\pi \frac{5^{-\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{-\cos x}} d x \\\\
& 2 \mathrm{I}=\int_0^\pi\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x \\\\
& {2 I}={2} \int_0^{\frac{\pi}{2}}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x
\end{aligned}
$$
$$ \begin{aligned} & I=\int_0^{\frac{\pi}{2}}\left(1+\sin x(-\sin 3 x)+\sin ^2 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 x+\cos ^3 x \cos 3 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}} 3+\cos 4 x+\left(\frac{\cos 3 x+3 \cos x}{4}\right) \cos 3 x-\sin 3 x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x \end{aligned} $$
$$ \begin{aligned} & 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 \mathrm{x}+\frac{1}{4}+\frac{3}{4} \cos 4 \mathrm{x}\right) \mathrm{dx} \\\\ & 2 \mathrm{I}=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4 \mathrm{x}}{4}\right)_0^{\frac{\pi}{2}} \Rightarrow \mathrm{I}=\frac{13 \pi}{16} \end{aligned} $$
$$ \begin{aligned} & I=\int_0^{\frac{\pi}{2}}\left(1+\sin x(-\sin 3 x)+\sin ^2 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 x+\cos ^3 x \cos 3 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}} 3+\cos 4 x+\left(\frac{\cos 3 x+3 \cos x}{4}\right) \cos 3 x-\sin 3 x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x \end{aligned} $$
$$ \begin{aligned} & 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 \mathrm{x}+\frac{1}{4}+\frac{3}{4} \cos 4 \mathrm{x}\right) \mathrm{dx} \\\\ & 2 \mathrm{I}=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4 \mathrm{x}}{4}\right)_0^{\frac{\pi}{2}} \Rightarrow \mathrm{I}=\frac{13 \pi}{16} \end{aligned} $$
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