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

解答

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

1 - cos (x)

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

= - [- cos (u)]_{x}^{0}

带入上下限计算后得:

- 1 + cos (x)

= - (- 1 + cos (x))

化简

= 1 - cos (x)

\int_{0}^{1} 3 x^{2} - 3 x d x

\int_{1}^{e} \frac{( 1 + ln ( x ) ) ^{4}}{x} d x

\int_{- 1}^{1} e^{3 x} d x

\int_{0}^{1} ln (1 + y) d y

\int_{0}^{2} 2 x (x^{2} + 1)^{3} d x