# How do you integrate int x^2sin^2x?

Nov 8, 2017

$\frac{{x}^{3}}{6} + \frac{x \cdot \cos 2 x}{4} - \left({x}^{2} - 2\right) \cdot \frac{\sin 2 x}{4} + C$

#### Explanation:

$\int {x}^{2} \cdot {\left(\sin x\right)}^{2} \cdot \mathrm{dx}$

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

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

After using tabular integration for second integral.

$\int {x}^{2} \cdot \cos 2 x \cdot \mathrm{dx}$

=${x}^{2} \cdot \frac{\sin 2 x}{2} - 2 x \cdot \frac{\cos 2 x}{- 4} + 2 \cdot \frac{\sin 2 x}{- 8}$

=$\left(2 {x}^{2} - 4\right) \cdot \frac{\sin 2 x}{4} + \frac{x \cdot \cos 2 x}{2}$

Thus,

$\int {x}^{2} \cdot {\left(\sin x\right)}^{2} \cdot \mathrm{dx}$

=$\frac{{x}^{3}}{6} + C$-$\frac{1}{2} \cdot \left[\left(2 {x}^{2} - 4\right) \cdot \frac{\sin 2 x}{4} + \frac{x \cdot \cos 2 x}{2}\right]$

=$\frac{{x}^{3}}{6} + \frac{x \cdot \cos 2 x}{4} - \left({x}^{2} - 2\right) \cdot \frac{\sin 2 x}{4} + C$