The value of the integral  ${\int }_0^{\frac{\pi }{2}}60\frac{\sin \left(6x\right)}{\sin x}dx$

is equal to

 

$$\begin{gathered} I=60{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\sin 6x-\sin 4x}{\sin x}+\frac{\sin 4x-\sin 2x}{\sin x}+\frac{\sin 2x}{\sin x}\right) dx \\ \Rightarrow I=60{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2 \cos 5x+2 \cos 3x+2 \cos x\right) dx \\ \Rightarrow I=60{{\overline{)\left(\frac{2}{5}\sin 5x+\frac{2}{3}\sin 3x+2 \sin x\right)}}_0}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \Rightarrow I=120\left(\frac{1}{5}-\frac{1}{3}+1\right)=120\left(\frac{3-5+15}{15}\right)=8\times 13=104 \end{gathered}$$