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

解答

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

- 2

求解步骤

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

提出常数:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} x cos (2 x) d x

使用换元积分法

= 4 \cdot \int_{0}^{π} \frac{u cos ( u )}{4} d u

提出常数:

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

= 4 \cdot \frac{1}{4} \cdot \int_{0}^{π} u cos (u) d u

使用分布积分法

= 4 \cdot \frac{1}{4} [u sin (u) - \int sin (u) d u]_{0}^{π}

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

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

化简

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

= [u sin (u) + cos (u)]_{0}^{π}

带入上下限计算后得:

- 2

= - 2

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