integral from 0 to x^2 of sin(2sqrt(t))

解

\int_{0}^{x^{2}} sin (2 t) d t

\frac{1}{2} sin (2 x) - x cos (2 x)

解答ステップ

\int_{0}^{x^{2}} sin (2 t) d t

u置換積分法を適用する

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

定数を除く:

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

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

部分積分を適用する

= 2 [- \frac{1}{2} u cos (2 u) - \int - \frac{1}{2} cos (2 u) d u]_{0}^{x^{2}}

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

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

簡素化

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

= 2 [- \frac{1}{2} u cos (2 u) + \frac{1}{4} sin (2 u)]_{0}^{x}

限度を計算する:

\frac{1}{4} sin (2 x) - \frac{1}{2} x cos (2 x)

= 2 (\frac{1}{4} sin (2 x) - \frac{1}{2} x cos (2 x))

簡素化

= \frac{1}{2} sin (2 x) - x cos (2 x)