integral from 0 to pi/2 of xcos^3(x)

解

\int_{0}^{\frac{π}{2}} x cos^{3} (x) d x

\frac{π}{3} - \frac{7}{9}

十進法表記

0.26941 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} x cos^{3} (x) d x

部分積分を適用する

= [x (sin (x) - \frac{1}{3} sin^{3} (x)) - \int \frac{1}{3} (3 sin (x) - sin^{3} (x)) d x]_{0}^{\frac{π}{2}}

\int \frac{1}{3} (3 sin (x) - sin^{3} (x)) d x = \frac{1}{9} (- 6 cos (x) - cos^{3} (x))

= [x (sin (x) - \frac{1}{3} sin^{3} (x)) - \frac{1}{9} (- 6 cos (x) - cos^{3} (x))]_{0}^{\frac{π}{2}}

簡素化

[x (sin (x) - \frac{1}{3} sin^{3} (x)) - \frac{1}{9} (- 6 cos (x) - cos^{3} (x))]_{0}^{\frac{π}{2}} : [\frac{1}{9} (9 x sin (x) - 3 x sin^{3} (x) + 6 cos (x) + cos^{3} (x))]_{0}^{\frac{π}{2}}

= [\frac{1}{9} (9 x sin (x) - 3 x sin^{3} (x) + 6 cos (x) + cos^{3} (x))]_{0}^{\frac{π}{2}}

限度を計算する:

\frac{π}{3} - \frac{7}{9}

= \frac{π}{3} - \frac{7}{9}