integral of usin^2(3u)

解

\int u sin^{2} (3 u) d u

\frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u)) + C

解答ステップ

\int u sin^{2} (3 u) d u

三角関数の公式を使用して書き換える

= \int \frac{1}{2} u (1 - cos (6 u)) d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u (1 - cos (6 u)) d u

部分積分を適用する

= \frac{1}{2} (u (u - \frac{1}{6} sin (6 u)) - \int u - \frac{1}{6} sin (6 u) d u)

\int u - \frac{1}{6} sin (6 u) d u = \frac{u ^{2}}{2} + \frac{1}{36} cos (6 u)

= \frac{1}{2} (u (u - \frac{1}{6} sin (6 u)) - (\frac{u ^{2}}{2} + \frac{1}{36} cos (6 u)))

簡素化

\frac{1}{2} (u (u - \frac{1}{6} sin (6 u)) - (\frac{u ^{2}}{2} + \frac{1}{36} cos (6 u))) : \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u))

= \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u))

定数を解答に追加する

= \frac{1}{72} (18 u^{2} - 6 u sin (6 u) - cos (6 u)) + C