Verified Solution • 2025–26 Session moderate

QJEEM23Ma16c9: JEE Main 2023 Mathematics - Class 12 JEEM23Ma16c9 Solution: If , then k is equal to _____________.

If ∫0π5cos⁡x(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)dx1+5cos⁡x=kπ16\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 _____________.

Step-by-Step Verified Solution

amp;I=∫0π5cos⁡x(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)1+5cos⁡xdxamp;I=∫0π5−cos⁡x(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)1+5−cos⁡xdxamp;2I=∫0π(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)dxamp;2I=2∫0π2(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)dx\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}

amp;I=∫0π2(1+sin⁡x(−sin⁡3x)+sin⁡2x−sin⁡3xsin⁡3x)dxamp;2I=∫0π2(3+cos⁡4x+cos⁡3xcos⁡3x−sin⁡3xsin⁡3x)dxamp;2I=∫0π23+cos⁡4x+(cos⁡3x+3cos⁡x4)cos⁡3x−sin⁡3x(3sin⁡x−sin⁡3x4)dx\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}

amp;2I=∫0π2(3+cos⁡4x+14+34cos⁡4x)dxamp;2I=134×π2+74(sin⁡4x4)0π2⇒I=13π16\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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