integral of x^2cos^2(x)

解

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

\frac{1}{3} x^{3} cos^{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^{2} cos^{2} (x) d x

部分積分を適用する

= \frac{1}{3} x^{3} cos^{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} cos^{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} cos^{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} cos^{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} cos^{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} cos^{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