Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go further :)

Note $f(x)=\sin^3 x \cos^2 2x$. You have
$$\cos 2x = 2 \cos^2 x -1$$
Hence
$$(\sin^3 x)(\cos^2 2x)=\sin x \sin^2 x (2 \cos^2 x -1)=\sin x (1- \cos^2 x) (2 \cos^2 x -1)$$
As $(\cos x)^\prime = -\sin x$, we get by expending the right hand side $$f(x)=-(\cos x)^\prime (- 2 \cos^4 x +3 \cos^2 x -1)$$ and finally
$$\int f(x)dx = \frac{2}{5} \cos^5 x - \cos^3 x + \cos x + a$$
