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.
 A: 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
$$2I=-\int_{0}^{\pi/2} \sin^2x \cos^2 x \sin 6x ~ dx=\frac{1}{4}\left(\int_{0}^{\pi/2} (\cos4x-1) \sin 6x~ dx \right).$$ $$ \Rightarrow I=\frac{1}{8}\left(\int_{0}^{\pi/2} \sin 6x  \cos 4x -\sin 6x ~dx \right)= \left (\int_{0}^{\pi/2} \frac{ \sin 10x +\sin 2x}{16}-\frac{\sin 6x}{8} \right) dx.$$ $$\Rightarrow I =\frac{1}{160}+\frac{1}{32}-\frac{1}{48}=\frac{1}{60}.$$
A: Hint: Use that $$\sin(x)^3\cos(x)^2\cos(7x)=\frac{1}{32} (\sin (2 x)-\sin (4 x)-2 \sin (6 x)+2 \sin (8 x)+\sin (10 x)-\sin (12 x))$$
A: Hint:
$$\sin3x=3\sin x-4\sin^3x,\cos2x=2\cos^2x-1$$
$$\implies(4\sin^3x)(2\cos^2x)=(3\sin x-\sin3x)(1+\cos2x)=3\sin x-\sin3x+3\sin x\cos2x-\sin2x\cos2x$$
Use Werner Formulas to express each term as $\sin(ax)$
Alternatively,
Use Intuition behind euler's formula,
$$2i\sin y=e^{iy}-e^{-iy},2\cos y=e^{iy}=e^{-iy},$$
$$(2i\sin x)^3(2\cos x)^2(2\cos7x)=(e^{ix}-e^{-ix})^3(e^{ix}+e^{-ix})^2(e^{i7x}+e^{-i7x})$$
$$\implies-i2^6\sin^3x\cos^2x\cos7x$$
$$=(e^{i3x}-3e^{ix}+3e^{-ix}-e^{-i3x})(e^{i2x}+e^{-i2x}+2)(e^{i7x}+e^{-i7x})$$
