JEE Main 2024 — 1 February, Shift 2 — Question 11

$\int_{0}^{\pi / 3} \cos ^{4} \mathrm{xdx}$

$=\int_{0}^{\pi / 3}\left(\frac{1+\cos 2 x}{2}\right)^{2} d x$

$=\frac{1}{4} \int_{0}^{\pi / 3}\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x$

$=\frac{1}{4}\left[\int_{0}^{\pi / 3} \mathrm{dx}+2 \int_{0}^{\pi / 3} \cos 2 \mathrm{xdx}+\int_{0}^{\pi / 3} \frac{1+\cos 4 \mathrm{x}}{2} \mathrm{dx}\right]$

$=\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 \mathrm{x})_{0}^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 \mathrm{x})_{0}^{\pi / 3}\right]$

$=\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 \mathrm{x})_{0}^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 \mathrm{x})_{0}^{\pi / 3}\right]$

$=\frac{1}{4}\left[\frac{\pi}{2}+\frac{\sqrt{3}}{2}+\frac{1}{8} \times\left(-\frac{\sqrt{3}}{2}\right)\right]$

$=\frac{\pi}{8}+\frac{7 \sqrt{3}}{64}$

$\therefore \mathrm{a}=\frac{1}{8} ; \mathrm{b}=\frac{7}{64}$

$\therefore 9 a+8 b=\frac{9}{8}+\frac{7}{8}=2$