Question Number 98744 by M±th+et+s last updated on 15/Jun/20
$$\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{2}{sin}\theta} \left(\mathrm{1}+{rsin}\theta\right){r}\:{dr}\:{d}\theta \\ $$
Commented by bemath last updated on 16/Jun/20
![= ∫_0 ^π [ (1/2)r^2 +(1/3)r^3 sin θ]_0 ^(2sin θ) dθ = ∫_0 ^π (2sin^2 θ+(8/3)sin^4 θ) dθ = ∫_0 ^π (1−cos 2θ+(8/3)((1/2)−(1/2)cos 2θ)^2 dθ = ∫_0 ^π (1−cos 2θ+(8/3)((1/4)−(1/2)cos 2θ+(1/4)((1/2)+(1/2)cos 4θ)) dθ = ∫_0 ^π (1−cos 2θ+1 −(4/3)cos 2θ+(1/3)cos 4θ ) dθ = [ 2θ−(7/6)sin 2θ +(1/(12))sin 4θ ]_0 ^π = 2π ■](https://www.tinkutara.com/question/Q98753.png)
$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left[\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\mathrm{r}^{\mathrm{3}} \mathrm{sin}\:\theta\right]_{\mathrm{0}} ^{\mathrm{2sin}\:\theta} \mathrm{d}\theta \\ $$$$=\:\overset{\pi} {\int}_{\mathrm{0}} \left(\mathrm{2sin}\:^{\mathrm{2}} \theta+\frac{\mathrm{8}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{4}} \theta\right)\:\mathrm{d}\theta \\ $$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta\right)^{\mathrm{2}} \:\mathrm{d}\theta\right. \\ $$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4}\theta\right)\right)\:\mathrm{d}\theta\right. \\ $$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}\:−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:\mathrm{4}\theta\:\right)\:\mathrm{d}\theta \\ $$$$=\:\left[\:\mathrm{2}\theta−\frac{\mathrm{7}}{\mathrm{6}}\mathrm{sin}\:\mathrm{2}\theta\:+\frac{\mathrm{1}}{\mathrm{12}}\mathrm{sin}\:\mathrm{4}\theta\:\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\:\mathrm{2}\pi\:\blacksquare \\ $$
Commented by M±th+et+s last updated on 16/Jun/20
$${thank}\:{you}\:{sir} \\ $$