Answer
$= \frac{1}{\beta}[-x^{2}cos(\beta x)) + (\frac{2x}{\beta}\sin (\beta x)) + \frac{2}{\beta^{2}}(\cos (\beta x)]+C$
Work Step by Step
$\int x^{2} \sin (\beta x) dx$
1. Find $du$
$u = x^{2}$
$u' = 2x$
$\frac{du}{dx} = 2x$
$du = 2x dx$
2. Find $v$
$dv = \sin (\beta x) $
$v = -\frac{1}{\beta}cos(\beta x)$
3. Substitute $u$ and $v$ into the equation
$uv - \int vdu$
$=(x^{2})(-\frac{1}{\beta}cos(\beta x)) - \int -\frac{1}{\beta}cos(\beta x) 2xdx$
$=(x^{2})(-\frac{1}{\beta}cos(\beta x)) - \int -\frac{2}{\beta}cos(\beta x) xdx$
$=(x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}\int cos(\beta x) xdx$
4. Find $\int cos(\beta x) xdx$ using by parts again. After that, you substitute that part back into Part 3.
$\int cos(\beta x) xdx$
4.1 Find $du$
$u = x$
$u' = 1$
$\frac{du}{dx} = 1$
$du = 1dx$
$du = dx$
4.2 Find $v$
$dv = \cos \beta xdx$
$v = \frac{1}{\beta}\sin (\beta x)$
4.3 Find $uv-\int vdu$, using components from only Part 4.
$= (x)(\frac{1}{\beta}\sin (\beta x)) - \int (\frac{1}{\beta}\sin (\beta x))dx$
$= (x)(\frac{1}{\beta}\sin (\beta x)) - \frac{1}{\beta}\int \sin (\beta x)dx$
$= (x)(\frac{1}{\beta}\sin (\beta x)) - \frac{1}{\beta}(-\cos (\beta x)(\frac{1}{\beta}))$
$= (x)(\frac{1}{\beta}\sin (\beta x)) + \frac{1}{\beta^{2}}(\cos (\beta x))$
3. (Continued) Take the answer from 4.3 and substitute it back into the original equation
$= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}\int cos(\beta x) xdx$
$= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}[(x)(\frac{1}{\beta}\sin (\beta x)) + \frac{1}{\beta^{2}}(\cos (\beta x))]$
$= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}[(\frac{x}{\beta}\sin (\beta x)) + \frac{1}{\beta^{2}}(\cos (\beta x))]$
$= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + (\frac{2x}{\beta^{2}}\sin (\beta x)) + \frac{2}{\beta^{3}}(\cos (\beta x))$
$= (-\frac{x^{2}}{\beta}cos(\beta x)) + (\frac{2x}{\beta^{2}}\sin (\beta x)) + \frac{2}{\beta^{3}}(\cos (\beta x))$
$= \frac{1}{\beta}[-x^{2}cos(\beta x)) + (\frac{2x}{\beta}\sin (\beta x)) + \frac{2}{\beta^{2}}(\cos (\beta x)]$
