1. We need to evaluate $I=\int_0^{\pi/2}60\,\frac{\sin(6x)}{\sin x}\,dx.$

So, $I=60\int_0^{\pi/2}\frac{\sin(6x)}{\sin x}\,dx.$

2. Use the standard identity $\frac{\sin(nx)}{\sin x}=2\cos x\,U_{n-1}(\cos x),$ but for small $n$ it is easier to expand directly. In particular, $\frac{\sin(6x)}{\sin x}=2\cos x+2\cos 3x+2\cos 5x.$

This follows from the known formula: $\frac{\sin(2mx)}{\sin x}=2\sum_{k=1}^{m}\cos((2k-1)x).$ For $m=3$, $\frac{\sin(6x)}{\sin x}=2(\cos x+\cos 3x+\cos 5x).$

3. Substitute into the integral: $I=60\int_0^{\pi/2}2(\cos x+\cos 3x+\cos 5x)\,dx$ $=120\left(\int_0^{\pi/2}\cos x\,dx+\int_0^{\pi/2}\cos 3x\,dx+\int_0^{\pi/2}\cos 5x\,dx\right).$

4. Evaluate each integral:

• $\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1.$

• $\int_0^{\pi/2}\cos 3x\,dx=\left[\frac{\sin 3x}{3}\right]_0^{\pi/2}=\frac{\sin(3\pi/2)}{3}=-\frac13.$

• $\int_0^{\pi/2}\cos 5x\,dx=\left[\frac{\sin 5x}{5}\right]_0^{\pi/2}=\frac{\sin(5\pi/2)}{5}=\frac15.$

So the sum is $1-\frac13+\frac15=\frac{15-5+3}{15}=\frac{13}{15}.$

5. Therefore, $I=120\cdot\frac{13}{15}=8\cdot 13=104.$

Hence, $\boxed{104}.$