integral from-pi to pi of z^2cos(2z)

解答

\int_{- π}^{π} z^{2} cos (2 z) d z

π

十进制

3.14159 \dots

求解步骤

\int_{- π}^{π} z^{2} cos (2 z) d z

使用分布积分法

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

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

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

化简

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

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

带入上下限计算后得:

π

= π

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