Indefinite integral of $\sin^8(x)$ Suppose we have the following function:
$$\sin^8(x)$$
We have to find its anti-derivative
To find the indefinite integral of $\sin^4(x)$, I converted everything to $\cos(2x)$ and $\cos(4x)$ and then  integrated. However this method wont be suitable to find the indefinite integral $\sin^8(x)$ since we have to expand a lot. Is there any other way I can evaluate it easily, and more efficiently?
 A: I copied and pasted this answer, since I actually wrote this answer to a different question which you can find here: Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? ).
Define ${S_n = \int\sin^{2n}(x)dx}$. Then
$${S_{n}=\int \sin^2(x)\sin^{2n-2}(x)dx=S_{n-1}-\int \cos^2(x)\sin^{2n-2}(x)dx}$$
On the rightmost integral, using integration by parts yields
$${\int\cos^2(x) \sin^{2n-2}(x)dx=\frac{\cos(x)\sin^{2n-1}(x)}{2n-1}+\frac{1}{2n-1}\int \sin^{2n}(x)dx}$$
So overall
$${\Rightarrow S_n = S_{n-1}-\frac{\cos(x)\sin^{2n-1}(x)}{2n-1} - \frac{1}{2n-1}S_n}$$
And so
$${\left(\frac{2n}{2n-1}\right)S_n = S_{n-1} - \frac{\cos(x)\sin^{2n-1}(x)}{2n-1}}$$
$${\Rightarrow S_n = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
Now you have a recursion relation that will help you compute the integral for higher even powers of ${\sin(x)}$:
$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
A: The development through the double angle formulae is not so long, let me show.
\begin{align}
\sin^8x
  &=(\sin^2x)^4=\\
  &=\left(\frac{1-\cos2x}{2}\right)^4=\\
  &=\frac{1}{16}(1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x)=\\
  &=\frac{1}{16}[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+(\cos^22x)^2]=\\
  &=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+\left(\frac{1+\cos4x}{2}\right)^2\right]=\\
  &=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+{}\right.\\
    &\qquad\qquad\qquad\left.+\frac{1}{4}(1+2\cos4x+\cos^24x)\right]=\\
  &=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)+\right.\\
    &\qquad\qquad\qquad\left.+\frac{1}{4}\left(1+2\cos4x+\frac{1+\cos8x}{2}\right)\right]=\\
\end{align}
so we have
\begin{align}
\int\sin^8xdx 
  &=\frac{1}{16}\left[x-2\sin2x+3\left(x+\frac{1}{4}\sin4x\right)-2\left(\sin2x-\frac{1}{3}\sin^32x\right)+{}\right.\\
  &\qquad\left.\frac{1}{4}\left(x+\frac{1}{2}\sin4x+\frac{1}{2}\left(x+\frac{1}{8}\sin8x\right)\right)\right]+C
\end{align}
A: There is well known recursion:
$$\boxed{I_n=\int\sin^nx \space dx = -\frac{1}{n}\sin^{n-1}x \cdot \cos x + \frac{n-1}{n} \cdot I_{n-2}}.$$
A: Hint: assume $  z=\cos x+i\sin x$ so $\frac{1}{z}=\cos x-i\sin x$ and $\sin x=\frac{1}{2i} (z-\frac{1}{z})$, also  $z^n =\cos nx+i\sin nx$
A: By expanding
\begin{align}
\sin^8x 
  &=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)=\\
  &=\frac{1}{128}\left(
      \frac{e^{8ix}-e^{-8ix}}{2}
     -8\frac{e^{6ix}-e^{-6ix}}{2}
    +28\frac{e^{4ix}-e^{-4ix}}{2}
    -56\frac{e^{2ix}-e^{-2ix}}{2}+35\right)=\\
  &=\frac{1}{128}\left[\cos8x-8\cos6x+28\cos4x-56\cos2x+35\right]
\end{align}
or by using the identity
$$
(\sin x)^{2m}=\frac{2}{4^m}\left[\sum_{k=0}^{m-1}\binom{2m}{k}(-1)^{m-k}\cos[2(m-k)x]+\frac{1}{2}\binom{2m}{m}\right]
$$
that for $m=4$ provides
\begin{align}
\sin^8x 
  &=\frac{2}{4^4}\left[\sum_{k=0}^3\binom{8}{k}(-1)^k\cos[2(4-k)x]+\frac{1}{2}\binom{8}{4}\right]=\\
  &=\frac{1}{128}\left[\cos8x-8\cos6x+28\cos4x-56\cos2x+35\right]
\end{align}
we have
\begin{align}
\int\sin^8xdx 
  &=\frac{1}{128}\left[\frac{1}{8}\sin8x-\frac{4}{3}\sin6x+7\sin4x-28\sin2x+35x\right]+C\\
\end{align}
