integral from 0 to pi of t^2cos(3t)

解答

\int_{0}^{π} t^{2} cos (3 t) d t

- \frac{2 π}{9}

十进制

- 0.69813 \dots

求解步骤

\int_{0}^{π} t^{2} cos (3 t) d t

使用换元积分法

= \int_{0}^{3 π} \frac{u ^{2} cos ( u )}{27} d u

提出常数:

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

= \frac{1}{27} \cdot \int_{0}^{3 π} u^{2} cos (u) d u

使用分布积分法

= \frac{1}{27} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3 π}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{27} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{3 π}

带入上下限计算后得:

- 6 π

= \frac{1}{27} (- 6 π)

化简

= - \frac{2 π}{9}

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