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 :)
 A: 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
$$\int f(x)dx = \frac{2}{5} \cos^5 x - \cos^3 x + \cos x + a$$
A: 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 $$\int\left( \sin^3 x-4\sin^5 x+4\sin^7 x \right)\, dx = -\int\left(1-u^2-4(1-u^2)^2+4(1-u^2)^3\right)\, du$$
A: *

*Odd powers are easy to integrate, substitution $u=\cos x$ will convert it into a polynomial.

*Half-angle substitutions (or straight-up the universal trig substitution) can handle the even cases.

*If the integral goes over a simple multiple of the period, there are special forms (involving Beta function, for instance).

*If you really don't know what to do, you can always go into complex numbers, where all these are just exponentials, which are trivial to integrate. A lot of terms to process, of course, but that way, you don't have to remember all the multiple-angle formulas.

A: The most elementary method is this: $\sin ^3 x = \sin x (1- \cos ^2 x)$ and $\cos ^2 2x = (2 \cos^2 x -1) ^2$. Substitute, do the multiplications, group together similar terms and split your result into integrals of the form $\int \sin x \cos ^n x \Bbb d x$ (preceded by various factors). Note now that $- \sin x = (\cos x)'$, so the previous integral becomes $- \int (\cos x)' \cos ^ n \Bbb d x = - \frac {\cos ^{n+1} x} {n+1}$. Add up all these integrals and that is it.
A: $$\int\left(\sin^3(x)\cos^2(2x)\right)dx=$$
$$\int \left(\frac{7}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{3}{16}\sin(5x)-\frac{1}{16}\sin(7x)\right)dx=$$
$$\int \frac{7}{16}\sin(x)dx-\int \frac{5}{16}\sin(3x)dx+\int \frac{3}{16}\sin(5x)dx-\int \frac{1}{16}\sin(7x)dx =$$
$$\frac{7}{16}\int \sin(x)dx-\frac{5}{16}\int \sin(3x)dx+\frac{3}{16}\int \sin(5x)dx-\frac{1}{16}\int \sin(7x)dx =$$
$$-\frac{7}{16}\cos(x)-\frac{5}{16}\left(-\frac{1}{3}\cos(3x)\right)+\frac{3}{16}\left(-\frac{1}{5}\cos(5x)\right)-\frac{1}{16}\left(-\frac{1}{7}\cos(7x)\right) =$$
$$-\frac{7}{16}\cos(x)+\frac{5}{48}\cos(3x)-\frac{3}{80}\cos(5x)+\frac{1}{112}\cos(7x)$$
