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Evaluate $\int \sin ^6 x d x$
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Use $\sin ^2 x=(1-\cos (2 x)) / 2$ to rewrite the function:
\begin{aligned} \int \sin ^6 x d x=\int\left(\sin ^2 x\right)^3 d x &=\int \frac{(1-\cos 2 x)^3}{8} d x \\ &=\frac{1}{8} \int 1-3 \cos 2 x+3 \cos ^2 2 x-\cos ^3 2 x d x . \end{aligned}
Now we have four integrals to evaluate:
$\int 1 d x=x$
and
$\int-3 \cos 2 x d x=-\frac{3}{2} \sin 2 x$
are easy. The $\cos ^3 2 x$ integral is like the previous example:
\begin{aligned} \int-\cos ^3 2 x d x &=\int-\cos 2 x \cos ^2 2 x d x \\ &=\int-\cos 2 x\left(1-\sin ^2 2 x\right) d x \\ &=\int-\frac{1}{2}\left(1-u^2\right) d u \\ &=-\frac{1}{2}\left(u-\frac{u^3}{3}\right) \\ &=-\frac{1}{2}\left(\sin 2 x-\frac{\sin ^3 2 x}{3}\right) . \end{aligned}
And finally we use another trigonometric identity, $\cos ^2 x=(1+\cos (2 x)) / 2$ :
$\int 3 \cos ^2 2 x d x=3 \int \frac{1+\cos 4 x}{2} d x=\frac{3}{2}\left(x+\frac{\sin 4 x}{4}\right) .$
So at long last we get
$\int \sin ^6 x d x=\frac{x}{8}-\frac{3}{16} \sin 2 x-\frac{1}{16}\left(\sin 2 x-\frac{\sin ^3 2 x}{3}\right)+\frac{3}{16}\left(x+\frac{\sin 4 x}{4}\right)+C .$
by Platinum (141,884 points)

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