求不定积分 ∫x*(cosx)^2 dx 要详细过程的 高手快来啊

2024-12-03 20:49:32
推荐回答(1个)
回答1:

∫xcos^2 x dx
=∫x(cos2x+1)/2 dx
=1/2*∫xcos2xdx+1/2*∫xdx
=1/4∫xcos2xd2x+1/4∫dx^2
=1/4∫xdsin2x +x^2/4
=1/4 *xsin2x-1/4∫sin2xdx +x^2/4
=xsin2x/4+x^2/4-1/8∫sin2xd2x
=xsin2x/4+x^2/4+1/8∫dcos2x
=xsin2x/4+x^2/4+cos2x/8 + C