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

解答

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

求解步骤

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

使用三角恒等式改写

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

乘开

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

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

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} (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))))

化简

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))

解答补常数

= \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

\frac{d}{d x} (\frac{2 x}{( x + 1 )})

\int \frac{2 x + 1}{x ^{2} + 2 x + 5} d x

\int \frac{1}{2} e^{2 x} d x

x y^{^{'}} + 3 y = x^{3} - x

\int sin^{9} (x) cos^{3} (x) d x