integral of cos^3(x)cos(3x)

解

\int cos^{3} (x) cos (3 x) d x

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

解答ステップ

\int cos^{3} (x) cos (3 x) d x

簡素化

cos^{3} (x) : cos^{2} (x) cos (x)

= \int cos^{2} (x) cos (x) cos (3 x) d x

部分積分を適用する

= \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) - \int - \frac{1}{2} sin (2 x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) d x

\int - \frac{1}{2} sin (2 x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) d x = - \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x)))

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

簡素化

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

定数を解答に追加する

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