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

解答

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

求解步骤

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

使用三角恒等式改写

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

乘开

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

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

使用积分加法定则:

\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))))

化简

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))

解答补常数

= \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

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