∫sin^4 3x dx 怎么求?1/8x-1/12sin2x+1/96sin4x+C

网友回答

∫sin^4 3xdx
=∫(1-cos6x)^2/4 dx
=1/4∫1+cos^2 6x-2cos6x)dx
=1/4(∫dx+∫(1+cos12x)/2 dx -2∫cos6xdx)
=1/4(x+1/2x+1/6sin12x-1/3cos6x)
=3/8x+1/24sin12x-1/12cos6x +C
======以下答案可供参考======
供参考答案1:
∫sin^4 3x dx
=∫（1-cos^2 3x)sin^2 3x dx
=∫sin^2 3x- cos^2 3x* sin^2 3x dx
=∫sin^2 3x- 1/2* sin^2 6x dx
=.....
∫sin²xdx=∫[(1-cos2x)/2]dx=(1/2)x-(1/2)∫cos2xdx=(1/2)x-(1/4)∫cos2xd(2x)=(1/2)x-(1/4)sin2x+C