trigonometry - Show that $int_{0}^{pi/2} sin^3x cos^2 x cos7x ~ dx= frac{1}{60}.$

Show that $$\int_{0}^{\pi/2} \sin^3x \cos^2 x \cos7x ~ dx= \frac{1}{60}$$
I could solve this integral by making use of the special expansion $$\cos 7x=64 \cos^7 x- 112\cos^5 x + 56 \cos^3 x -7\cos x$$ and then using $\cos x =t.$ But I would like to know if there is a simpler way to solve this integral.

Answer

Use $$\int_{0}^{a} f(x) dx=\int_{0}^a f(a-x) dx. ~~~(1)$$ Then $$I=\int_{0}^{\pi/2} \sin^3x \cos^2 x \cos7x ~ dx. ~~~(2)$$ becomes $$I=\int_{0}^{\pi/2} \cos^3x \sin^2 x (-\sin7x) ~ dx. ~~~(3)$$ By adding (2) and (3), we get
$$2I=-\int_{0}^{\pi/2} \sin^2x \cos^2 x \sin 6x ~ dx=\frac{1}{4}\left(\int_{0}^{\pi/2} (\cos4x-1) \sin 6x~ dx \right).$$ $$\Rightarrow I=\frac{1}{8}\left(\int_{0}^{\pi/2} \sin 6x \cos 4x -\sin 6x ~dx \right)= \left (\int_{0}^{\pi/2} \frac{ \sin 10x +\sin 2x}{16}-\frac{\sin 6x}{8} \right) dx.$$ $$\Rightarrow I =\frac{1}{160}+\frac{1}{32}-\frac{1}{48}=\frac{1}{60}.$$