# How do you evaluate the integral int cos^3xsin^3x?

Jan 4, 2017

$= \frac{1}{4} {\sin}^{4} x - \frac{1}{6} {\sin}^{6} x + C$

#### Explanation:

We want to rewrite so that either sine or cosine is all by itself with a power of $1$ and use a u-substitution to eliminate it, leaving all terms in u that are easier to integrate.

$= \int \cos x \left({\cos}^{2} x {\sin}^{3} x\right)$

Rewrite ${\cos}^{2} x$ using the pythagorean identity ${\sin}^{2} x + {\cos}^{2} x = 1$.

$= \int \cos x \left(1 - {\sin}^{2} x\right) {\sin}^{3} x \mathrm{dx}$

$= \int \cos x \left({\sin}^{3} x - {\sin}^{5} x\right) \mathrm{dx}$

The derivative of $\sin x$ is $\cos x$. Accordingly, we take the substitution $u = \sin x$. Then $\mathrm{du} = \cos x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{\cos} x$

$= \int \cos x \left({u}^{3} - {u}^{5}\right) \cdot \frac{\mathrm{du}}{\cos} x$

This will eliminate, leaving us only with u's.

$= \int {u}^{3} - {u}^{5} \mathrm{du}$

Integrate using ${x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$.

$= \frac{1}{4} {u}^{4} - \frac{1}{6} {u}^{6} + C$

Resubstitute:

$= \frac{1}{4} {\left(\sin x\right)}^{4} - \frac{1}{6} {\left(\sin x\right)}^{6} + C$

$= \frac{1}{4} {\sin}^{4} x - \frac{1}{6} {\sin}^{6} x + C$

Hopefully this helps!

Jan 4, 2017

The answer is $= \frac{1}{192} \cos 6 x - \frac{3}{64} \cos 2 x + C$

#### Explanation:

${\cos}^{3} x = \frac{1}{4} \left(\cos 3 x + 3 \cos x\right)$

${\sin}^{3} x = \frac{1}{4} \left(3 \sin x - \sin 3 x\right)$

${\cos}^{3} x \cdot {\sin}^{3} x = \frac{1}{16} \left(\cos 3 x + 3 \cos x\right) \left(3 \sin x - \sin 3 x\right)$

$= \frac{1}{16} \left(3 \cos 3 x \sin x - \cos 3 x \sin 3 x + 9 \cos x \sin x - 3 \sin 3 x \cos x\right)$

$9 \cos x \sin x = \frac{9}{2} \sin 2 x$

$\cos 3 x \sin 3 x = \frac{1}{2} \sin 6 x$

$- 3 \sin 3 x \cos x + 3 \cos 3 x \sin x = - 3 \left(\sin \left(3 x - x\right)\right) = - 3 \sin 2 x$

Therefore,

${\cos}^{3} x \cdot {\sin}^{3} x = \frac{1}{16} \left(\frac{9}{2} \sin 2 x - 3 \sin 2 x - \frac{1}{2} \sin 6 x\right)$

$= \frac{3}{32} \sin 2 x - \frac{1}{32} \sin 6 x$

So,

$\int {\cos}^{3} x \cdot {\sin}^{3} x \mathrm{dx} = \frac{3}{32} \int \sin 2 x \mathrm{dx} - \frac{1}{32} \int \sin 6 x \mathrm{dx}$

$= \left(\frac{3}{32} \cdot - \cos 2 x \cdot \frac{1}{2}\right) - \left(\frac{1}{32} \cdot - \cos 6 x \cdot \frac{1}{6}\right) + C$

$= \frac{1}{192} \cos 6 x - \frac{3}{64} \cos 2 x + C$