integral from 0 to pi/2 of (x+1)cos(2x)

解答

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

- \frac{1}{2}

十进制

- 0.5

求解步骤

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

使用分布积分法

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

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

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

化简

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

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

带入上下限计算后得:

- \frac{1}{2}

= - \frac{1}{2}

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