Question

求 ∫ x(cosx)^2dx

是不是把(cosx)^2换成(cos2x+1)/2呢 然后怎么做啊

Answer 1

∫ xcos²x dx
= (1/2)∫ x(1+cos2x) dx
= (1/2)∫ xdx + (1/2)∫ xcos2x dx
= (1/2)(x²/2) + (1/4)∫ xcos2x d(2x)
= x²/4 + (1/4)∫ xd[∫ cos2x d(2x)] <= 这步可以不写，只因好让你明白
= x²/4 + (1/4)∫ xd(sin2x)
= x²/4 + (1/4)[xsin2x - ∫ sin2x dx]，这里运用分部积分法：∫ udv = uv - ∫ vdu，u是比v复杂的函数
= x²/4 + (1/4)xsin2x - (1/8)∫ sin2x d(2x)
= x²/4 + (1/4)xsin2x - (1/8)(-cos2x) + C
= x²/4 + (1/4)xsin2x + (1/8)cos2x + C

Answer 2

(用f代替积分号哈。。。>_<)然后=(1/2)fxcos2xdx+（1/2）fxdx 前面用分部积分法=(1/4)fxd(sin2x)=(1/4)xsin2x-(1/4)sin2xdx=(1/2)xcos2x-(1/4)sin2x 后面是(1/4)x^2 所以最后=（1/4）xsin2x+(1/8)cos2x+(1/4)x^2 +c