## Calculus: Early Transcendentals 8th Edition

$= \frac{1}{\beta}[-x^{2}cos(\beta x)) + (\frac{2x}{\beta}\sin (\beta x)) + \frac{2}{\beta^{2}}(\cos (\beta x)]+C$
$\int x^{2} \sin (\beta x) dx$ 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 $= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}\int cos(\beta x) xdx$ $= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}[(x)(\frac{1}{\beta}\sin (\beta x)) + \frac{1}{\beta^{2}}(\cos (\beta x))]$ $= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + \frac{2}{\beta}[(\frac{x}{\beta}\sin (\beta x)) + \frac{1}{\beta^{2}}(\cos (\beta x))]$ $= (x^{2})(-\frac{1}{\beta}cos(\beta x)) + (\frac{2x}{\beta^{2}}\sin (\beta x)) + \frac{2}{\beta^{3}}(\cos (\beta x))$ $= (-\frac{x^{2}}{\beta}cos(\beta x)) + (\frac{2x}{\beta^{2}}\sin (\beta x)) + \frac{2}{\beta^{3}}(\cos (\beta x))$ $= \frac{1}{\beta}[-x^{2}cos(\beta x)) + (\frac{2x}{\beta}\sin (\beta x)) + \frac{2}{\beta^{2}}(\cos (\beta x)]$