Correct option is (1) 2
\(\int\limits_{0}^{\pi / 3} \cos ^{4} x d x\)
\(=\int\limits_{0}^{\pi / 3}\left(\frac{1+\cos 2 x}{2}\right)^{2} d x\)
\(=\frac{1}{4} \int\limits_{0}^{\pi / 3}\left(1+2 \cos 2 x+\cos ^{2} 2 x\right) d x\)
\(=\frac{1}{4}\left[\int\limits_{0}^{\pi / 3} {dx}+2 \int\limits_{0}^{\pi / 3} \cos 2 {xdx}+\int\limits_{0}^{\pi / 3} \frac{1+\cos 4 {x}}{2} {dx}\right]\)
\(=\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_{0}^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_{0}^{\pi / 3}\right]\)
\(=\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_{0}^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_{0}^{\pi / 3}\right]\)
\(=\frac{1}{4}\left[\frac{\pi}{2}+\frac{\sqrt{3}}{2}+\frac{1}{8} \times\left(-\frac{\sqrt{3}}{2}\right)\right]\)
\(=\frac{\pi}{2}+\frac{7 \sqrt{3}}{64}\)
\(\therefore \mathrm{a}=\frac{1}{8} ; \mathrm{b}=\frac{7}{64}\)
\(\therefore 9 \mathrm{a}+8 \mathrm{~b}=\frac{9}{8}+\frac{7}{8}=2\)
