integral of (sin(3x)+3cos(3x))^2

解

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

5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x) + C

解答ステップ

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

拡張

(sin (3 x) + 3 cos (3 x))^{2} : sin^{2} (3 x) + 6 sin (3 x) cos (3 x) + 9 cos^{2} (3 x)

= \int sin^{2} (3 x) + 6 sin (3 x) cos (3 x) + 9 cos^{2} (3 x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sin^{2} (3 x) d x + \int 6 sin (3 x) cos (3 x) d x + \int 9 cos^{2} (3 x) d x

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

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

\int 9 cos^{2} (3 x) d x = \frac{9}{2} (x + \frac{1}{6} sin (6 x))

= \frac{1}{2} (x - \frac{1}{6} sin (6 x)) - cos^{2} (3 x) + \frac{9}{2} (x + \frac{1}{6} sin (6 x))

簡素化

\frac{1}{2} (x - \frac{1}{6} sin (6 x)) - cos^{2} (3 x) + \frac{9}{2} (x + \frac{1}{6} sin (6 x)) : 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x)

= 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x)

定数を解答に追加する

= 5 x - \frac{1}{12} sin (6 x) - cos^{2} (3 x) + \frac{3}{4} sin (6 x) + C