JEE Main 2026 — 6 April, Morning Shift — Question 38

Apply (P-5) $\mathrm{I}=\int_{-\pi / 4}^{\pi / 4} \frac{32 \cos ^{4} \theta}{1+\mathrm{e}^{-\mathrm{sin} \theta}} \mathrm{d} \theta$ Add $2 \mathrm{I}=\int_{-\pi / 4}^{\pi / 4} 32 \cos ^{4} \theta \mathrm{~d} \theta=2 \int_{0}^{\pi / 4} 32 \cos ^{4} \theta \mathrm{~d} \theta$ $I=32 \int_{0}^{\pi / 4} \cos ^{4} \theta d \theta$ $=8 \int_{0}^{\pi / 4}\left(2 \cos ^{2} \theta\right)^{2} \mathrm{~d} \theta=8 \int_{0}^{\pi / 4}(1+\cos 2 \theta)^{2} \mathrm{~d} \theta$ $=8 \int_{0}^{\pi / 4} 1+2 \cos 2 \theta+\cos ^{2} 2 \theta \mathrm{~d} \theta$ $=8 \int_{0}^{\pi / 4} 1+2 \cos 2 \theta+\frac{1+\cos 4 \theta}{2} d \theta$ $=8\left[\frac{3 \theta}{2}+\frac{2 \sin 2 \theta}{2}+\frac{\sin 4 \theta}{8}\right]_{0}^{\pi / 4}$ $=8\left[\left(\frac{3 \pi}{8}+1+0\right)\right]=3 \pi+8$