If ∫ _ 0 ^ (π)/(3) ^ 4 xdx = aπ + b 3 , where a and b are rational...

Let, $I = {∫}_{0}^{\dfrac{\pi}{3}}\cos^{4}x d x$

$⇒ I = \dfrac{1}{4}{∫}_{0}^{\dfrac{\pi}{3}}{\left(2\cos^{2}x\right)}^{2} d x$

$⇒ I = \dfrac{1}{4}{∫}_{0}^{\dfrac{\pi}{3}}{\left(1 + \cos 2x\right)}^{2} d x$

$⇒ I = \dfrac{1}{4}{∫}_{0}^{\dfrac{\pi}{3}}\left(1 + \cos^{2}2x + 2\cos 2x\right) d x$

$⇒ I = \dfrac{1}{4}{∫}_{0}^{\dfrac{\pi}{3}}\left(1 + 2\cos 2x\right) d x + \dfrac{1}{8}{∫}_{0}^{\dfrac{\pi}{3}}2\cos^{2}2xdx$

$⇒ I = \dfrac{1}{4}{∫}_{0}^{\dfrac{\pi}{3}}\left(1 + 2\cos 2x\right) d x + \dfrac{1}{8}{∫}_{0}^{\dfrac{\pi}{3}}\left(1 + \cos 4x\right)dx$

$⇒ I = {\left[\dfrac{1}{4}\left(x + \sin 2x\right) + \dfrac{1}{8}\left(x + \dfrac{\sin 4x}{4}\right)\right]}_{0}^{\dfrac{\pi}{3}}$

$⇒ I = \dfrac{1}{4}\left(\dfrac{\pi}{3} + sin\dfrac{2\pi}{3}\right) + \dfrac{1}{8}\left(\dfrac{\pi}{3} + \dfrac{sin\dfrac{4\pi}{3}}{4}\right)$

$⇒ I = \left(\dfrac{\pi}{12} + \dfrac{\sqrt{3}}{8}\right) + \left(\dfrac{\pi}{24} - \dfrac{\sqrt{3}}{64}\right)$

$⇒ I = \dfrac{\pi}{8} + \dfrac{7\sqrt{3}}{64}$

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

$⇒ 9a + 8b = \dfrac{9}{8} + \dfrac{7}{8}$

$⇒ 9a + 8b = \dfrac{16}{8}$

$⇒ 9a + 8b = 2$