integral from 0 to pi of sin(mx)

解答

\int_{0}^{π} sin (m x) d x

\frac{- cos ( πm ) + 1}{m}

求解步骤

\int_{0}^{π} sin (m x) d x

使用换元积分法

= \int_{0}^{mπ} \frac{sin ( u )}{m} d u

提出常数:

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

= \frac{1}{m} \cdot \int_{0}^{mπ} sin (u) d u

使用积分常见定则:

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

= \frac{1}{m} [- cos (u)]_{0}^{mπ}

带入上下限计算后得:

- cos (πm) + 1

= \frac{1}{m} (- cos (πm) + 1)

化简

= \frac{- cos ( πm ) + 1}{m}

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