integral of (xsin(x))^2

解

\int (x sin (x))^{2} d x

\frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x) + C

解答ステップ

\int (x sin (x))^{2} d x

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= \int x^{2} sin^{2} (x) d x

部分積分を適用する

= \frac{1}{3} x^{3} sin^{2} (x) - \int \frac{1}{3} x^{3} sin (2 x) d x

\int \frac{1}{3} x^{3} sin (2 x) d x = - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= \frac{1}{3} x^{3} sin^{2} (x) - (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))

簡素化

\frac{1}{3} x^{3} sin^{2} (x) - (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))) : \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)

= \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)

定数を解答に追加する

= \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x) + C