integral from 0 to x of 2tcos(t)

解

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

2 (x sin (x) + cos (x) - 1)

解答ステップ

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

定数を除く:

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

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

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

x sin (x) + cos (x) - 1

= 2 (x sin (x) + cos (x) - 1)