integral of cos(3x)(2+sin(3x))^4

解答

\int cos (3 x) (2 + sin (3 x))^{4} d x

\frac{16}{3} sin (3 x) - \frac{16}{3} cos^{2} (3 x) + \frac{8}{3} sin^{3} (3 x) + \frac{2}{3} sin^{4} (3 x) + \frac{1}{15} sin^{5} (3 x) + C

求解步骤

\int cos (3 x) (2 + sin (3 x))^{4} d x

乘开

cos (3 x) (2 + sin (3 x))^{4} : 16 cos (3 x) + 32 cos (3 x) sin (3 x) + 24 sin^{2} (3 x) cos (3 x) + 8 sin^{3} (3 x) cos (3 x) + sin^{4} (3 x) cos (3 x)

= \int 16 cos (3 x) + 32 cos (3 x) sin (3 x) + 24 sin^{2} (3 x) cos (3 x) + 8 sin^{3} (3 x) cos (3 x) + sin^{4} (3 x) cos (3 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 16 cos (3 x) d x + \int 32 cos (3 x) sin (3 x) d x + \int 24 sin^{2} (3 x) cos (3 x) d x + \int 8 sin^{3} (3 x) cos (3 x) d x + \int sin^{4} (3 x) cos (3 x) d x

\int 16 cos (3 x) d x = \frac{16}{3} sin (3 x)

\int 32 cos (3 x) sin (3 x) d x = - \frac{16}{3} cos^{2} (3 x)

\int 24 sin^{2} (3 x) cos (3 x) d x = \frac{8}{3} sin^{3} (3 x)

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

\int sin^{4} (3 x) cos (3 x) d x = \frac{1}{15} sin^{5} (3 x)

= \frac{16}{3} sin (3 x) - \frac{16}{3} cos^{2} (3 x) + \frac{8}{3} sin^{3} (3 x) + \frac{2}{3} sin^{4} (3 x) + \frac{1}{15} sin^{5} (3 x)

解答补常数

= \frac{16}{3} sin (3 x) - \frac{16}{3} cos^{2} (3 x) + \frac{8}{3} sin^{3} (3 x) + \frac{2}{3} sin^{4} (3 x) + \frac{1}{15} sin^{5} (3 x) + C

\frac{d}{d a} (sin (a) + sin (2 a))

\frac{d}{d x} (\frac{23}{x ^{4}})

\int \frac{1}{tan ( 5 θ )} d θ

f (x) = x^{\frac{1}{2}}

\int_{1}^{2} ln (x - 1) d x