integral from 0 to pi/4 of xsin(2x)

解答

\int_{0}^{\frac{π}{4}} x sin (2 x) d x

\frac{1}{4}

十进制

0.25

求解步骤

\int_{0}^{\frac{π}{4}} x sin (2 x) d x

使用分布积分法

= [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{\frac{π}{4}}

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{4}}

化简

[\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{4}} : [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{4}}

带入上下限计算后得:

\frac{1}{4}

= \frac{1}{4}

\int_{0}^{9} (x - \frac{1}{3} x) d x

\int e^{2 x} cos (7 x) d x

z \to 1 - i lim (x + i (2 x + y))

\frac{\partial}{\partial x} (sin (π (2 x - y)))

\int x^{7} d x