integral of sin^2(y)cos^4(y)

解

\int sin^{2} (y) cos^{4} (y) d y

\frac{3}{8} y - \frac{3}{32} sin (4 y) + \frac{19}{24} cos^{3} (y) sin (y) - \frac{1}{6} cos^{5} (y) sin (y) - \frac{5}{32} (2 y + sin (2 y)) + C

解答ステップ

\int sin^{2} (y) cos^{4} (y) d y

三角関数の公式を使用して書き換える

= \int (1 - cos^{2} (y)) cos^{4} (y) d y

拡張

(1 - cos^{2} (y)) cos^{4} (y) : cos^{4} (y) - cos^{6} (y)

= \int cos^{4} (y) - cos^{6} (y) d y

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 cos^{4} (y) d y - \int cos^{6} (y) d y

\int cos^{4} (y) d y = cos^{3} (y) sin (y) + \frac{3}{8} (y - \frac{1}{4} sin (4 y))

\int cos^{6} (y) d y = \frac{sin ( y ) cos ^{5} ( y )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (y) sin (y) + \frac{3}{8} (y + \frac{1}{2} sin (2 y)))

= cos^{3} (y) sin (y) + \frac{3}{8} (y - \frac{1}{4} sin (4 y)) - (\frac{sin ( y ) cos ^{5} ( y )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (y) sin (y) + \frac{3}{8} (y + \frac{1}{2} sin (2 y))))

簡素化

cos^{3} (y) sin (y) + \frac{3}{8} (y - \frac{1}{4} sin (4 y)) - (\frac{sin ( y ) cos ^{5} ( y )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (y) sin (y) + \frac{3}{8} (y + \frac{1}{2} sin (2 y)))) : \frac{3}{8} y - \frac{3}{32} sin (4 y) + \frac{19}{24} cos^{3} (y) sin (y) - \frac{1}{6} cos^{5} (y) sin (y) - \frac{5}{32} (2 y + sin (2 y))

= \frac{3}{8} y - \frac{3}{32} sin (4 y) + \frac{19}{24} cos^{3} (y) sin (y) - \frac{1}{6} cos^{5} (y) sin (y) - \frac{5}{32} (2 y + sin (2 y))

定数を解答に追加する

= \frac{3}{8} y - \frac{3}{32} sin (4 y) + \frac{19}{24} cos^{3} (y) sin (y) - \frac{1}{6} cos^{5} (y) sin (y) - \frac{5}{32} (2 y + sin (2 y)) + C