求不定积分∫xcos2xdx 求过程答案!谢谢!!!

2024-12-21 21:23:55
推荐回答(3个)
回答1:

∫ xcos2x dx
= (1/2)∫ xcos2x d2x
= (1/2)∫ x dsin2x
= (1/2)xsin2x - (1/2)∫ sin2x dx
= (1/2)xsin2x - (1/2)(1/2)(-cos2x) + C
= (1/2)xsin2x + (1/4)cos2x + C

回答2:

∫xcos2xdx =∫x2cos2xdx/2 =(xsin2x-∫sin2xdx)/2=(xsinx-∫2sin2xdx/2)/2=xsinx/2+cos2x/4

回答3:

4sinx^2-2x^2