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

解答

\int_{0}^{π} x^{2} cos (x + 1) d x

π^{2} sin (π + 1) - 2 (- π cos (π + 1) + sin (π + 1)) + 2 sin (1)

十进制

- 8.33392 \dots

求解步骤

\int_{0}^{π} x^{2} cos (x + 1) d x

使用分布积分法

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

\int 2 x sin (x + 1) d x = 2 (- x cos (x + 1) + sin (x + 1))

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

带入上下限计算后得:

π^{2} sin (π + 1) - 2 (- π cos (π + 1) + sin (π + 1)) + 2 sin (1)

= π^{2} sin (π + 1) - 2 (- π cos (π + 1) + sin (π + 1)) + 2 sin (1)

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