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?

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}
