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 :)

take your integrand $$\sin^3 x-4\sin^5 x+4\sin^7 x $$ and factor the $\sin x$ so you get $$ \sin x\left(\sin^2 x-4\sin^4 x+4\sin^6 x\right) = \sin x\left( (1-\cos^2 x)-4(1-\cos^2 x)^2+4(1-\cos^2 x)^3  \right)$$
now make the substitution $$u = \cos x , \quad du = -\sin x \,dx $$ so you have $$\int\left( \sin^3 x-4\sin^5 x+4\sin^7 x \right)\, dx = -\int\left(1-u^2-4(1-u^2)^2+4(1-u^2)^3\right)\, du$$
