# What is the integral of int sin^2(x)cos^4(x) dx?

Jun 5, 2016

$\setminus \frac{1}{16} \setminus \left(x - \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x \setminus\right)\right) + \setminus \frac{1}{6} \setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x\right) + C$

#### Explanation:

$\setminus \int \setminus {\sin}^{2} \setminus \left(x \setminus\right) {\cos}^{4} \setminus \left(x \setminus\right) \mathrm{dx}$

Applying integral reduction,

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

so,
$\setminus \int \setminus {\sin}^{2} \setminus \left(x \setminus\right) \setminus {\cos}^{4} \setminus \left(x \setminus\right) \mathrm{dx}$
$= \setminus \frac{\setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x \setminus\right)}{6} + \setminus \frac{3}{6} \setminus \int \setminus {\cos}^{2} \setminus \left(x \setminus\right) \setminus {\sin}^{2} \setminus \left(x \setminus\right) \mathrm{dx}$

$= \setminus \frac{\setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x \setminus\right)}{6} + \setminus \frac{3}{6} \setminus \int \setminus {\cos}^{2} \setminus \left(x \setminus\right) \setminus {\sin}^{2} \setminus \left(x \setminus\right) \mathrm{dx}$

We know,
$\setminus \int \setminus {\cos}^{2} \setminus \left(x \setminus\right) \setminus {\sin}^{2} \setminus \left(x \setminus\right) \mathrm{dx} = \setminus \frac{1}{8} \setminus \left(x - \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x \setminus\right) \setminus\right)$

Then,
$= \setminus \frac{\setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x \setminus\right)}{6} + \setminus \frac{3}{6} \setminus \frac{1}{8} \setminus \left(x - \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x \setminus\right) \setminus\right)$

Simplifying,
$= \setminus \frac{1}{16} \setminus \left(x - \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x \setminus\right) \setminus\right) + \setminus \frac{1}{6} \setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x \setminus\right)$

$= \setminus \frac{1}{16} \setminus \left(x - \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x \setminus\right) \setminus\right) + \setminus \frac{1}{6} \setminus {\sin}^{3} \setminus \left(x \setminus\right) \setminus {\cos}^{3} \setminus \left(x \setminus\right) + C$