integral from 0 to x^6 of cos(sqrt(t))

解答

\int_{0}^{x^{6}} cos (t) d t

2 (x^{3} sin (x^{3}) + cos (x^{3}) - 1)

求解步骤

\int_{0}^{x^{6}} cos (t) d t

使用换元积分法

= \int_{0}^{x^{6}} cos (u) \cdot 2 u d u

提出常数:

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

= 2 \cdot \int_{0}^{x^{6}} cos (u) u d u

使用分布积分法

= 2 [u sin (u) - \int sin (u) d u]_{0}^{x^{6}}

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

= 2 [u sin (u) - (- cos (u))]_{0}^{x^{6}}

化简

2 [u sin (u) - (- cos (u))]_{0}^{x^{6}} : 2 [u sin (u) + cos (u)]_{0}^{x^{3}}

= 2 [u sin (u) + cos (u)]_{0}^{x^{3}}

带入上下限计算后得:

x^{3} sin (x^{3}) + cos (x^{3}) - 1

= 2 (x^{3} sin (x^{3}) + cos (x^{3}) - 1)

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