Evaluar la integral de este ejercicio

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¡Hola Francisco!

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$$\begin{align}&\int_0^{\pi/2}\int_{0}^{sen\, \theta}\theta r\;dr\,d\theta=\\&\\&\int_0^{\pi/2}\left[ \theta· \frac{r^2}{2}\right]_0^{sen \,\theta}\;d\theta=\\&\\&\int_0^{\pi/2} \left(\frac{\theta}2 sen^2 \theta  \right)d\theta=\\&\\&\int_0^{\pi/2} \left(\frac{\theta}2·\frac{1-\cos 2\theta}{2} \right)d\theta=\\&\\&\int_0^{\pi/2} \left(\frac{\theta}{4}-\frac{\theta \cos 2\theta}{4}  \right)d\theta=\\&\\&\left[\frac {\theta^2}{8}\right]_0^{\pi/2}-\frac 14\int_0^{\pi/2}\theta \cos 2\theta\;d\theta=\\&\\&u=\theta\qquad\qquad\qquad du=d\theta\\&dv= \cos 2\theta\;d\theta \qquad \;v=\frac{sen \,2\theta}2\\&\\&\frac{\pi^2}{32}-\left[\frac{\theta\,sen\,2\theta}{8}\right]_0^{\pi/2}+\frac 18\int_0^{\pi/2}sen\,2\theta\;d\theta=\\&\\&\frac{\pi^2}{32}-0+0-\left[\frac 1{16}\cos 2\theta\right]_0^{\pi/2}\\&\\&\frac{\pi^2}{32}-\left(-\frac 1{16}-\frac 1{16}  \right)=\\&\\&\frac{\pi^2}{32}+\frac 18=\frac{\pi^2+4}{32}\end{align}$$

Y eso es todo, espero que te siva y lo hayas entendido.

Saludos

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