integral from 0 to t of cos^2(wt)

解

\int_{0}^{t} cos^{2} (wt) d t

\frac{1}{2} (t + \frac{sin ( 2 wt )}{2 w})

解答ステップ

\int_{0}^{t} cos^{2} (wt) d t

三角関数の公式を使用して書き換える

= \int_{0}^{t} \frac{1 + cos ( 2 wt )}{2} d t

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{t} 1 + cos (2 wt) d t

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_{0}^{t} 1 d t + \int_{0}^{t} cos (2 wt) d t)

\int_{0}^{t} 1 d t = t

\int_{0}^{t} cos (2 wt) d t = \frac{sin ( 2 wt )}{2 w}

= \frac{1}{2} (t + \frac{sin ( 2 wt )}{2 w})