# Calc1.61

$\int \sin ^{5}x\cos ^{5}x\,dx\,$

Trigonometric integrals are split up into a few cases. This problem is the case where either the $\sin x$ or $\cos x$ is to an odd power. Both are odd, in fact, for this problem but the procedure is the same if only one is odd. First, pick the one that has an odd power, either will work here. For this specific example, let's turn 4 of those $\cos x$ into $\sin x$ and what we'll have left is a whole bunch of $\sin x$ and 1 $\cos x$, which is the derivative of $\sin x$. This sets up a nice integral by substitution.

$I=\int \sin ^{5}x\cos ^{5}x\,dx=\int \sin ^{5}x(1-\sin ^{2}x)^{2}\cos x\,dx=\int (\sin ^{9}x-2\sin ^{7}x+\sin ^{5}x)\cos x\,dx\,$

Let $u=\sin x\,$ and $du=\cos x\,dx\,$

$I=\int (u^{9}-2u^{7}+u^{5})\,du={\frac {1}{10}}u^{{10}}-{\frac {1}{4}}u^{8}+{\frac {1}{6}}u^{6}+C\,$

And substitution back in gives us

$\int \sin ^{5}x\cos ^{5}x\,dx={\frac {1}{10}}\sin ^{{10}}x-{\frac {1}{4}}\sin ^{8}x+{\frac {1}{6}}\sin ^{6}x+C\,$

The procedure for many integrals is the same with the main difference being which of the trigonometric substitutions are used and in which way they are used.