integral from 0 to pi of tsin(2t)

解答

\int_{0}^{π} t sin (2 t) d t

- \frac{π}{2}

十进制

- 1.57079 \dots

求解步骤

\int_{0}^{π} t sin (2 t) d t

使用分布积分法

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

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

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

化简

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

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

带入上下限计算后得:

- \frac{π}{2}

= - \frac{π}{2}

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