7.1 Integration By Parts/7
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Revision as of 05:56, 29 November 2022 by Kimberlyr70044@students.laalliance.org (talk | contribs) (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...")
<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>