Kimberlyr70044@students.laalliance.org: Created page with " $\begin{align} &=\int x^2 sin (\pi x)\\[2ex] & U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex] &U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex] &= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x..."$

2022-11-29T05:57:44Z

Created page with "<math> \begin{align} &=\int x^2 sin (\pi x)\\[2ex] & U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex] &U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex] &= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x..."

New page

<math>
\begin{align}
&=\int x^2 sin (\pi x)\\[2ex]
& U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
&U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex]
&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
\end{align}
</math>

7.1 Integration By Parts/38 - Revision history