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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

簡素化

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

三角関数の公式を使用して書き換える

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

sum from n=1 to infinity of-1

f(x)=4sin(2x)-3cos(3x)

integral from 1/5 to 7 of 14xln(5x)