Integration Questions

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

$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left[\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\mathrm{r}^{\mathrm{3}} \mathrm{sin}\:\theta\right]_{\mathrm{0}} ^{\mathrm{2sin}\:\theta} \mathrm{d}\theta \\$$$$=\:\overset{\pi} {\int}_{\mathrm{0}} \left(\mathrm{2sin}\:^{\mathrm{2}} \theta+\frac{\mathrm{8}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{4}} \theta\right)\:\mathrm{d}\theta \\$$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta\right)^{\mathrm{2}} \:\mathrm{d}\theta\right. \\$$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4}\theta\right)\right)\:\mathrm{d}\theta\right. \\$$$$=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}\:−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:\mathrm{4}\theta\:\right)\:\mathrm{d}\theta \\$$$$=\:\left[\:\mathrm{2}\theta−\frac{\mathrm{7}}{\mathrm{6}}\mathrm{sin}\:\mathrm{2}\theta\:+\frac{\mathrm{1}}{\mathrm{12}}\mathrm{sin}\:\mathrm{4}\theta\:\right]_{\mathrm{0}} ^{\pi} \\$$$$=\:\mathrm{2}\pi\:\blacksquare \\$$

Commented by  M±th+et+s last updated on 16/Jun/20

$${thank}\:{you}\:{sir} \\$$