Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}60\frac{\sin 6x}{\sin x}dx$

$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}60\frac{2\sin 3x\cos 3x}{\sin x}dx$

$\Rightarrow I=60{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\left(3\sin x-4{\sin }^{3}x\right)\left(4{\cos }^{3}x-3\cos x\right)}{\sin x}dx$

$\Rightarrow I=120{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(3-4{\sin }^{2}x\right)\left(4{\cos }^{2}x-3\right)\cos xdx$

$\Rightarrow I=120{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(3-4{\sin }^{2}x\right)\left(1-4{\sin }^{2}x\right)\cos xdx$

Now let $\sin x=t\Rightarrow \cos xdx=dt$ and limit changes to $0 \text{to }1$,

So, $I=120{\int }_0^1\left(3-4t^2\right)\left(1-4t^2\right)dt$

$\Rightarrow I=120{\int }_0^{1}3-16t^2+16t^4 dt$

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

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