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

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

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

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

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

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

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

$$ = 104$$