0-pi-0-2sin-1-rsin-r-dr-d-

Question Number 98744 by M±th+et+s last updated on 15/Jun/20

\int_{0}^{π} \int_{0}^{2 s i n θ} (1 + r s i n θ) r d r d θ

Commented by bemath last updated on 16/Jun/20

= \underset{0}{\int^{π}} {[\frac{1}{2} r^{2} + \frac{1}{3} r^{3} \sin θ]}_{0}^{2 \sin θ} d θ

= {\int^{π}}_{0} (2 \sin^{2} θ + \frac{8}{3} \sin^{4} θ) d θ

= \underset{0}{\int^{π}} (1 - \cos 2 θ + \frac{8}{3} {(\frac{1}{2} - \frac{1}{2} \cos 2 θ)}^{2} d θ

= \underset{0}{\int^{π}} (1 - \cos 2 θ + \frac{8}{3} (\frac{1}{4} - \frac{1}{2} \cos 2 θ + \frac{1}{4} (\frac{1}{2} + \frac{1}{2} \cos 4 θ)) d θ

= \underset{0}{\int^{π}} (1 - \cos 2 θ + 1 - \frac{4}{3} \cos 2 θ + \frac{1}{3} \cos 4 θ) d θ

= {[2 θ - \frac{7}{6} \sin 2 θ + \frac{1}{12} \sin 4 θ]}_{0}^{π}

= 2 π

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

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