integral from 0 to t of sin(w(t-x))

解答

\int_{0}^{t} sin (w (t - x)) d x

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

求解步骤

\int_{0}^{t} sin (w (t - x)) d x

使用换元积分法

= \int_{t}^{0} - sin (w u) d u

提出常数:

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

= - \int_{t}^{0} sin (w u) d u

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

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

带入上下限计算后得:

- 1 + cos (wt)

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

化简

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

\int_{1}^{2} \frac{1}{( x - 2 ) ^{\frac{1}{3}}} d x

\int_{0}^{2} 1 + 36 x^{2} d x

\int_{0}^{3} \frac{8 x}{3} d x

\int_{0}^{2} 16 - x^{2} - 2 y^{2} d x

\int_{0}^{19} x 19 - x d x