A Polar Double Integral

  ` \int_0^{\pi/2}\int_0^{1+\sin\theta}\theta\text{  }r\text{ d}r\text{ d}\theta `
= ` \int_0^{\pi/2}\theta[\frac{1}{2}r^2]_0^{1+\sin\theta}\text{ d}\theta `
= ` \int_0^{\pi/2}\frac{1}{2}\theta(1+\sin\theta)^2\text{ d}\theta `
= ` \frac{1}{2}\int_0^{\pi/2}\theta(1+2\sin\theta+\sin^2\theta)\text{ d}\theta `
= ` \frac{1}{2}\int_0^{\pi/2}\theta(1+2\sin\theta+\frac{1-cos2\theta}{2})\text{ d}\theta `
= ` \frac{1}{2}\int_0^{\pi/2}\theta(\frac{3}{2}+2\sin\theta-\frac{1}{2}cos2\theta)\text{ d}\theta `

= ` \frac{3}{4}\int_0^{\pi/2}\theta\text{ d}\theta + \int_0^{\pi/2}\theta(\sin\theta-\frac{1}{4}cos2\theta)\text{ d}\theta `
= ` \frac{3}{4}[\frac{\theta^2}{2}]_0^{\pi/2} - \int_0^{\pi/2}\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta+\frac{1}{8}sin2\theta)\text{ d}\theta `
= ` \frac{3}{8}\cdot\frac{\pi^2}{4} - [\theta(\cos\theta+\frac{1}{8}sin2\theta)]_0^{\pi/2} + \int_0^{\pi/2}\cos\theta+\frac{1}{8}sin2\theta\text{ d}\theta `
= ` \frac{3\pi^2}{32} - 0 + [\sin\theta-\frac{1}{16}cos2\theta]_0^{\pi/2} `
= ` \frac{3\pi^2}{32} + [(1+\frac{1}{16})-(0-\frac{1}{16})] `
= ` \frac{3\pi^2}{32} + \frac{9}{8} `