题目
求不定积分根号4-x^2/x^2dx
提问时间:2020-10-05
答案
令x=2sinu,则:sinu=x/2,u=arcsin(x/2),dx=(1/2)cosudu.
∴∫[√(4-x^2)/x^2]dx
=∫[cosu/(sinu)^2]cosudu
=∫[(cosu)^2/(sinu)^2]du
=∫{[1-(sinu)^2]/(sinu)^2}du
=∫[1/(sinu)^2]du-∫du
=-cotu-u+C
=-cosu/sinu-arcsin(x/2)+C
=-√[1-(sinu)^2]/(x/2)-arcsin(x/2)+C
=-√[1-(x/2)^2]/(x/2)-arcsin(x/2)+C
=-√(4-x^2)/x-arcsin(x/2)+C.
∴∫[√(4-x^2)/x^2]dx
=∫[cosu/(sinu)^2]cosudu
=∫[(cosu)^2/(sinu)^2]du
=∫{[1-(sinu)^2]/(sinu)^2}du
=∫[1/(sinu)^2]du-∫du
=-cotu-u+C
=-cosu/sinu-arcsin(x/2)+C
=-√[1-(sinu)^2]/(x/2)-arcsin(x/2)+C
=-√[1-(x/2)^2]/(x/2)-arcsin(x/2)+C
=-√(4-x^2)/x-arcsin(x/2)+C.
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