Question

∫cos^4xsin^2xdx怎么积分？

Answer 1

∵cos^4xsin^2x
=cos^4x(1-cos²x)
=cos^4x-cos^6x
=[1+cos(2x)]²/4-[1+cos(2x)]³/8
=[1+2cos(2x)+cos²(2x)]/4-[1+3cos(2x)+3cos²(2x)+cos³(2x)]/8
=1/8+1/8cos(2x)-1/8cos²(2x)-1/8cos³(2x)
=1/8+1/8cos(2x)-[1+cos(4x)]/16-1/8cos³(2x)
=1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)
∴∫cos^4xsin^2xdx
=∫[1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)]dx
=x/16+sin(2x)/16-sin(4x)/64-1/16∫[1-sin²(2x)]d[sin(2x)]
=x/16+sin(2x)/16-sin(4x)/64-[sin(2x)-sin³(2x)/3]/16+C
=x/16-sin(4x)/64+sin³(2x)/48+C, (C是积分常数)。

Answer 2

把cos四次降成二次cos2x
在凑微分就可以了

自己算下

Answer 3

变为cos^4x-cos^6x,这个分开积即可

Answer 4

你范围是什么吖?!

原式=|4(-sinx)^3*sin^2x+cos^4x*2*cosx|(...)