$6\int_{0}^{\pi/2} (\sin 3x + \sin 2x + \sin x) dx$ is equal to ______

$6\int_{0}^{\pi/2} 2\sin 2x \cos x + \sin 2x , dx$

$= 6\int_{0}^{\pi/2} (4\sin x \cos^2 x + 2\sin x \cos x) dx$

$I = 12\int_{0}^{\pi/2} \sin x (2\cos^2 x + \cos x) dx$

Put $\cos x = t \Rightarrow -\sin x dx = dt$

$I = -12\int_{1}^{0} (2t^2 + t) dt$

$I = 12\int_{0}^{1} (2t^2 + t) dt$

$I = 12\left[\frac{2t^3}{3} + \frac{t^2}{2}\right]_0^1 = 12\left(\frac{2}{3} + \frac{1}{2}\right) = 17$