求不定积分

Question

∫x³√（4-x²）dx 求不定积分。注：“√”是根号，根号不会打了就用这个代替了根号下（4-x²）。

Answer 1

用换元法：x=2cost
∫x³√(4-x²)dx=∫8cos^3t√(4-4cos^2t)d(2cost)=-32∫cos^3t*sin^2tdt=-32∫cos^2t*sin^2td(sint)
=-32∫sin^2td(sint)-32∫sin^4td(sint)=(-32/3)(sin^3t)-(32/5)(sin^5t)+c
又有sint=√(1-(x/2)^2)
只要代入上式即可
有不懂欢迎追问

Answer 2

yes or no

Answer 3

∫x^3√(4-x^2)dx
=(1/4)∫√(4-x^2)dx^4
=(1/2)∫x^2 *√(4-x^2) dx^2
=(1/2)∫(x^2-4)*√(4-x^2)dx^2 +(1/2)∫4√(4-x^2)dx^2
=(1/2)∫√(4-x^2)^3d(4-x^2)-2∫√(4-x^2)d(4-x^2)
=(1/5)√(4-x^2)^5 -(4/3)√(4-x^2)^3 +C本回答被提问者采纳