Solving Trig Substitution Homework: ∫√(4-x^2)/x dx

[neshepard]

Homework Statement

∫√(4-x^2)/x dx

Homework Equations

The Attempt at a Solution

a^2=4 u^2=x^2 ⇒ u=asinθ
a=2 u=x
x=2sinθ sinθ=x/2 (Our professor uses a triangle method which I won't draw)
2cosθ=√(4-x^2)
dx=2cosθ dθ

∫√(4-x^2)/x dx=∫2cosθ/2sinθ dθ
=∫cosθ/sinθ dθ
u=sinθ
du=cosθ dθ
=∫1/u du
=ln|u| + C
=ln|sinθ| + C

⇒resubstitute x for θ
=ln|x/2| + C

Is this correct?

 

[Mark44]

In a quick scan I don't see anything obviously wrong. You can check your answer by differentiating what you ended with. Its derivative should be sqrt(4 - x^2)/x.

 

[neshepard]

I tried that, but the problem is I get 1/x. If I check with my calculator I get:
2ln|[√(4-x^2)-2)/|x|] + √(4-x^2).

 

[Mark44]

∫√(4-x^2)/x dx=∫2cosθ/2sinθ dθ
=∫cosθ/sinθ dθ

There's a mistake above. You should get ∫2cosθ/2sinθ * 2cosθ dθ. You seem to have forgotten to substitute for dx.