求不定积分根号4-x^2/x^2dx
1个回答

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.