integral from 0 to t of sin(wt)

解答

\int_{0}^{t} sin (wt) d t

\frac{- cos ( wt ) + 1}{w}

求解步骤

\int_{0}^{t} sin (wt) d t

使用换元积分法

= \int_{0}^{wt} \frac{sin ( u )}{w} d u

提出常数:

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

= \frac{1}{w} \cdot \int_{0}^{wt} sin (u) d u

使用积分常见定则:

\int sin (u) d u = - cos (u)

= \frac{1}{w} [- cos (u)]_{0}^{wt}

带入上下限计算后得:

- cos (wt) + 1

= \frac{1}{w} (- cos (wt) + 1)

化简

= \frac{- cos ( wt ) + 1}{w}

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