The value of the integral ∫ _ 0 ^ (π)/(2) 60( (6x))/( x)dx is equal...

Let $I = {∫}_{0}^{\dfrac{\pi}{2}}60\dfrac{\sin 6x}{\sin x}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}60\dfrac{2\sin 3xcos3x}{\sin x}dx$

$⇒ I = 60{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2\left(3\sin x - 4\sin^{3}x\right)\left(4\cos^{3}x - 3\cos x\right)}{\sin x}dx$

$⇒ I = 120{∫}_{0}^{\dfrac{\pi}{2}}\left(3 - 4\sin^{2}x\right)\left(4\cos^{2}x - 3\right)cosxdx$

$⇒ I = 120{∫}_{0}^{\dfrac{\pi}{2}}\left(3 - 4\sin^{2}x\right)\left(1 - 4\sin^{2}x\right)cosxdx$

Now let $\sin x = t ⇒ cosxdx = dt$ and limit changes to $0\text{to}1$,

So, $I = 120{∫}_{0}^{1}\left(3 - 4{t}^{2}\right)\left(1 - 4{t}^{2}\right)dt$

$⇒ I = 120{∫}_{0}^{1}3 - 16{t}^{2} + 16{t}^{4}dt$

$⇒ I = 120{\left[3t - \dfrac{16{t}^{3}}{3} + \dfrac{16{t}^{5}}{5}\right]}_{0}^{1}$

$⇒ I = 120\left[3 - \dfrac{16}{3} + \dfrac{16}{5}\right] = 120 \times \dfrac{45 - 80 + 48}{15} = 8 \times 13 = 104$