integral of (2cos(x/2)-3cos(x/3))^2

解

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

\frac{13}{2} x - \frac{36}{5} sin (\frac{5 x}{6}) + \frac{27}{4} sin (\frac{2 x}{3}) + 2 sin (x) - 36 sin (\frac{x}{6}) + C

解答ステップ

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

拡張

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

= \int 4 cos^{2} (\frac{x}{2}) - 12 cos (\frac{x}{2}) cos (\frac{x}{3}) + 9 cos^{2} (\frac{x}{3}) 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 4 cos^{2} (\frac{x}{2}) d x - \int 12 cos (\frac{x}{2}) cos (\frac{x}{3}) d x + \int 9 cos^{2} (\frac{x}{3}) d x

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

\int 12 cos (\frac{x}{2}) cos (\frac{x}{3}) d x = 6 (\frac{6}{5} sin (\frac{5 x}{6}) + 6 sin (\frac{x}{6}))

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

= 2 (x + sin (x)) - 6 (\frac{6}{5} sin (\frac{5 x}{6}) + 6 sin (\frac{x}{6})) + \frac{9}{2} (x + \frac{3}{2} sin (\frac{2 x}{3}))

簡素化

2 (x + sin (x)) - 6 (\frac{6}{5} sin (\frac{5 x}{6}) + 6 sin (\frac{x}{6})) + \frac{9}{2} (x + \frac{3}{2} sin (\frac{2 x}{3})) : \frac{13}{2} x - \frac{36}{5} sin (\frac{5 x}{6}) + \frac{27}{4} sin (\frac{2 x}{3}) + 2 sin (x) - 36 sin (\frac{x}{6})

= \frac{13}{2} x - \frac{36}{5} sin (\frac{5 x}{6}) + \frac{27}{4} sin (\frac{2 x}{3}) + 2 sin (x) - 36 sin (\frac{x}{6})

定数を解答に追加する

= \frac{13}{2} x - \frac{36}{5} sin (\frac{5 x}{6}) + \frac{27}{4} sin (\frac{2 x}{3}) + 2 sin (x) - 36 sin (\frac{x}{6}) + C