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

解

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

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

解答ステップ

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

三角関数の公式を使用して書き換える

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

拡張

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

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

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} (t) d t - \int cos^{6} (t) d t

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

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

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

簡素化

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

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

定数を解答に追加する

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