integral of θsin^2(θ)

解

\int θ sin^{2} (θ) d θ

\frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ)) + C

解答ステップ

\int θ sin^{2} (θ) d θ

部分積分を適用する

= \frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \int \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) d θ

\int \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) d θ = \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ))

= \frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ))

定数を解答に追加する

= \frac{1}{2} θ (θ - \frac{1}{2} sin (2 θ)) - \frac{1}{2} (\frac{θ ^{2}}{2} + \frac{1}{4} cos (2 θ)) + C