Answer to Question #101354 in Calculus for Sanjaya

Question

Double Integral

1. Integral from 0 to pi, Integral from 0 to y^2 of sin x/y dxdy

Answer 1

$\int_{0}^{\pi} \int_{0}^{y^2} \sin\frac{x}{y}dxdy=\int_{0}^{\pi} dy\int_{0}^{y^2} \sin\frac{x}{y}dx=\\ \\ =\int_{0}^{\pi}\left(-y\cos\frac{x}{y}\right)\mid^{y^2}_{0} dy=\\ \\ =-\int_{0}^{\pi}\left(y\cos\frac{y^2}{y} - y\cos0\right)dy=\\ \\=-\int_{0}^{\pi}y\cos y dy +\int_{0}^{\pi}y dy=\\ \\ =-\left( y\sin y \mid^{\pi}_{0} - \int_{0}^{\pi}\sin y dy\right) + \frac{y^2}{2}\mid^{\pi}_{0}=\\ \\ =-\left(\pi\sin \pi - 0 + \cos y \mid^{\pi}_{0}\right) + \frac{\pi^2}{2}=\\ \\ =-\cos\pi + \cos 0 +\frac{\pi^2}{2} = 2+\frac{\pi^2}{2}$