If ∫ 0 π 3 cos 4 x d x = a π + b 3 , where a and b are rational numbers, then 9 a + 8 b is equal to:

Let,  $I={\int }_0^{\frac{\mathrm{\pi }}{3}}{\cos }^{4}xdx$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{3}}{\left(2{\cos }^{2}x\right)}^{2}dx$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{3}}{\left(1+\cos 2x\right)}^{2}dx$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{3}}\left(1+{\cos }^{2}2x+2\cos 2x\right)dx$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{3}}\left(1+2\cos 2x\right)dx+\frac{1}{8}{\int }_0^{\frac{\mathrm{\pi }}{3}}2{\cos }^{2}2xdx$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\frac{\mathrm{\pi }}{3}}\left(1+2\cos 2x\right)dx+\frac{1}{8}{\int }_0^{\frac{\mathrm{\pi }}{3}}\left(1+\cos 4x\right)dx$

$\Rightarrow I={\left[\frac{1}{4}\left(x+\sin 2x\right)+\frac{1}{8}\left(x+\frac{\sin 4x}{4}\right)\right]}_0^{\frac{\mathrm{\pi }}{3}}$

$\Rightarrow I=\frac{1}{4}\left(\frac{\mathrm{\pi }}{3}+\sin \frac{2\mathrm{\pi }}{3}\right)+\frac{1}{8}\left(\frac{\mathrm{\pi }}{3}+\frac{\sin \frac{4\mathrm{\pi }}{3}}{4}\right)$

$\Rightarrow I=\left(\frac{\mathrm{\pi }}{12}+\frac{\sqrt{3}}{8}\right)+\left(\frac{\mathrm{\pi }}{24}-\frac{\sqrt{3}}{64}\right)$

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

So, on comparing with given value we get, $a=\frac{1}{8}, b=\frac{7}{64}$

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

$\Rightarrow 9a+8b=\frac{16}{8}$

$\Rightarrow 9a+8b=2$