integral from 0 to 1 of xcos(ax)

解

\int_{0}^{1} x cos (a x) d x

\frac{a sin ( a ) + cos ( a ) - 1}{a ^{2}}

解答ステップ

\int_{0}^{1} x cos (a x) d x

u置換積分法を適用する

= \int_{0}^{a} \frac{u cos ( u )}{a ^{2}} d u

定数を除く:

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

= \frac{1}{a ^{2}} \cdot \int_{0}^{a} u cos (u) d u

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

a sin (a) + cos (a) - 1

= \frac{1}{a ^{2}} (a sin (a) + cos (a) - 1)

簡素化

= \frac{a sin ( a ) + cos ( a ) - 1}{a ^{2}}