積分法(定積分)急

∫√(x^2 - 4)/x dx [2,4]Let x = 2secθ, dx = 2secθtanθdθx = 2, θ = (sec1)^(-1)x = 4, θ = (sec2)^(-1)√(x^2 - 4)= √[4(secθ)^2 - 4]

= 2tanθ∫√(x^2 - 4)/x dx [2,4]= ∫ (2tanθ/2secθ)(2secθtanθ)dθ [(sec1)^(-1), (sec2)^(-1)]= 2 ∫ (tanθ)^2 dθ= 2 ∫ (secθ)^2 - 1 dθ= 2(tanθ - θ)= 2[(tan π/3 - π/3) - (tan 0 - 0)]= 2(√3 - π/3)

你係味要好像大約金做....
∫ √(x^2-r^2) dx
=Area of the 1/4 circle (with the radius r)
= [(π)r^2]/4

仲有無其他方法？如不用三角學跟設u...

Technique of Trigo is needed to solve