∫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
∫sin^4 3x dx 怎么求?
∫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