Question

∫x^3*cosx^2dx 求过程阿！感谢！

Answer 1

I=(1/2)∫ (x^2) cos(x^2) d(x^2)
令x^2=u,先是第一换元法,再用分部积分法,
I=(1/2)∫u*cosudu
=(1/2)[ud(sinu)]
=(1/2) [ u*sinu-∫sinudu]
=(1/2)[usinu+cosu+C)
=(1/2)[x^2sin(x^2)+cos(x^2)+C].

Answer 2

∫ x³cos²x dx
= ∫ x³(1 + cos2x)/2 dx
= ∫ x³/2 dx + (1/2)∫ x³cos2x dx
= x⁴/8 + (1/2)∫ x³ d((sin2x)/2)
= x⁴/8 + (1/4)x³sin2x - (1/4)∫ 3x²sin2x dx
= x⁴/8 + (1/4)x³sin2x - (3/4)∫ x² d(- (cos2x)/2)
= x⁴/8 + (1/4)x³sin2x + (3/8)x²cos2x - (3/8)∫ 2xcos2x dx
= x⁴/8 + (1/4)x³sin2x + (3/8)x²cos2x - (3/4)∫ x d((sin2x)/2)
= x⁴/8 + (1/4)x³sin2x + (3/8)x²cos2x - (3/8)xsin2x - (3/8)∫ sin2x dx
= x⁴/8 + (1/4)x³sin2x + (3/8)x²cos2x - (3/8)xsin2x + (3/8)(- (cos2x)/2) + C
= x⁴/8 + (1/4)x³sin(2x) + (3/8)x²cos(2x) - (3/8)xsin(2x) - (3/16)cos(2x) + C

追问

正确答案显示1／2（x^2sinx^2+cosx^2）+c。。但不知道怎么出来的

追答

sorry,2次方在x上。
∫ x³cos(x²) dx
= ∫ x²cos(x²) d(x²/2)
= (1/2)∫ x² d(sin(x²))
= (1/2)x²sin(x²) - (1/2)∫ sin(x²) d(x²)
= (1/2)x²sin(x²) + (1/2)cos(x²) + C
= (1/2)[x²sin(x²) + cos(x²)] + C