integral of t(1-t^2)^{1/2}

解

\int t (1 - t^{2})^{\frac{1}{2}} d t

\frac{1}{2} t^{2} (1 - t^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - t^{2})^{\frac{3}{2}} - 3 1 - t^{2}) + C

解答ステップ

\int t (1 - t^{2})^{\frac{1}{2}} d t

部分積分を適用する

= \frac{1}{2} t^{2} (1 - t^{2})^{\frac{1}{2}} - \int - \frac{t ^{3}}{2 ( 1 - t ^{2} ) ^{\frac{1}{2}}} d t

\int - \frac{t ^{3}}{2 ( 1 - t ^{2} ) ^{\frac{1}{2}}} d t = - \frac{1}{6} ((1 - t^{2})^{\frac{3}{2}} - 3 1 - t^{2})

= \frac{1}{2} t^{2} (1 - t^{2})^{\frac{1}{2}} - (- \frac{1}{6} ((1 - t^{2})^{\frac{3}{2}} - 3 1 - t^{2}))

簡素化

= \frac{1}{2} t^{2} (1 - t^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - t^{2})^{\frac{3}{2}} - 3 1 - t^{2})

定数を解答に追加する

= \frac{1}{2} t^{2} (1 - t^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - t^{2})^{\frac{3}{2}} - 3 1 - t^{2}) + C