Question

求解高数。。。

求解高数

Answer 1

弧长=∫(√3,√8) √(1+y'^2)dx

=∫(√3,√8) √(1+1/x^2)dx
=∫(√3,√8) √(x^2+1)/xdx
令x=tant，则dx=sec^2tdt
弧长=∫(arctan√3,arctan√8) sect/tant*sec^2tdt
=∫(arctan√3,arctan√8) 1/sintcos^2tdt
=∫(arctan√3,arctan√8) (sin^2t+cos^2t)/sintcos^2tdt
=∫(arctan√3,arctan√8) (tant*sect+csct)dt
=[sect+ln|csct-cott|]|(arctan√3,arctan√8)
=3+ln|3/√8-1/√8|-2-ln|2/√3-1/√3|
=1-(1/2)*ln2+(1/2)*ln3
=1+(1/2)*ln(3/2)
=1+ln(√6/2)