7.1 Integration By Parts/7: Difference between revisions

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

Revision as of 00:18, 30 November 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. TeX parse error: Bracket argument to \\ must be a dimension"): {\displaystyle {\begin{aligned}&=\int x^{2}sin(\pi x)\\[2ex]U=x^{2},\\[x]du=2xdx,\\[x]dv=sin(\pi x)dx,\\[x]v=-{\frac {1}{\pi }}cos(\pi x)\\[2ex]&=-{\frac {x^{2}}{\pi }}cos(\pi x)+{\frac {2}{\pi }}\int xcos(\pi x)dx\\[2ex]U=x,\\[x]du=dx,\\[x]dv=cos(\pi x)dx,\\[x]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{aligned}}}