Question

微积分用换元法计算定积分

Answer 1

let
x= secu
dx= secu.tanu du
x=-2, u=2π/3
x=-1, u=π

∫(-2->-1) √(x^2-1) /x
=∫(2π/3->π) (tanu)^2 du
=∫(2π/3->π) [ (secu)^2 -1] du
=[ tanu - u ](2π/3->π)
=-(π/3) - √3