"The value of the integral $\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to _________."

Question

Accepted Answer

"

$I = \int\limits_0^{{\pi \over 2}} {60\,.\,{{\sin 6x} \over {\sin x}}dx}$

$$ = 60\,.\,2\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(4{{\cos }^2}x - 3)\cos x\,dx} $$

$$ = 120\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(1 - 4{{\sin }^2}x)\cos x\,dx} $$

Let $\sin x = t \Rightarrow \cos xdx = dt$

$= 120\int\limits_0^1 {(3 - 4{t^2})(1 - 4{t^2})dt}$

$= 120\int\limits_0^1 {(3 - 16{t^2} + 16{t^4})dt}$

$= 120\left[ {3t - {{16{t^3}} \over 3} + {{16{t^5}} \over 5}} \right]_0^1$

$= 104$

"

The value of the integral $\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to _________.