integral from 0 to t of ycos(t-x)

解答

\int_{0}^{t} y cos (t - x) d x

y sin (t)

求解步骤

\int_{0}^{t} y cos (t - x) d x

提出常数:

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

= y \cdot \int_{0}^{t} cos (t - x) d x

使用换元积分法

= y \cdot \int_{t}^{0} - cos (u) d u

提出常数:

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

= y (- \int_{t}^{0} cos (u) d u)

使用积分常见定则:

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

= y (- [sin (u)]_{t}^{0})

化简

= - y [sin (u)]_{t}^{0}

带入上下限计算后得:

- sin (t)

= - y (- sin (t))

化简

= y sin (t)

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