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

解

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

\frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x)) + C

解答ステップ

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

三角関数の公式を使用して書き換える

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

拡張

sin^{4} (x) (1 - sin^{2} (x)) : sin^{4} (x) - sin^{6} (x)

= \int sin^{4} (x) - sin^{6} (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^{4} (x) d x - \int sin^{6} (x) d x

\int sin^{4} (x) d x = - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))

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

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

簡素化

- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) : \frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x))

= \frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x))

定数を解答に追加する

= \frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x)) + C