求不定积分∫cos²xdx

Question

如题所述

Answer 1

cos²x =(1 +cos 2x)/2
∫ cos²x dx =(1/2)∫(1 +cos 2x) dx
= (1/2)x +(1/4)sin 2x +C
∫cos²xdx =∫cosxdsinx=sinxcosx-∫sinxdcosx=sinxcosx+∫sin²xdx=sinxcosx+∫(1-cos²x)dx
=sinxcosx+x-∫cos²xdx
2∫cos²xdx =sinxcosx+x
∫cos²xdx =(sinxcosx+x)/2 +C (这个是分部积分法)和上面的答案结果是一样的(sinxcosx+x)/2=(sinxcosx)/2+x/2=(sin 2x)/4+ x/2

Answer 2

∫cos²xdx =∫cosxdsinx=sinxcosx-∫sinxdcosx=sinxcosx+∫sin²xdx=sinxcosx+∫(1-cos²x)dx
=sinxcosx+x-∫cos²xdx
2∫cos²xdx =sinxcosx+x
∫cos²xdx =(sinxcosx+x)/2 +C本回答被网友采纳

Answer 3

原式=∫(1+cos2x)/2dx
=1/2x+1/4sin2x+C

Answer 4

解：∫ cos²x dx =(1/2) ∫ (1 +cos 2x) dx
= x/2 +(1/4)sin 2x +C. (C为任意常数).

= = = = = = = = =
注意：∫ cos ax dx = (1/a) sin ax , (a不等于0).

Answer 5

sin²x