integral of (1-cos(2u))/2

解

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

\frac{1}{2} (u - \frac{1}{2} sin (2 u)) + C

解答ステップ

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

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

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

定数を解答に追加する

= \frac{1}{2} (u - \frac{1}{2} sin (2 u)) + C