Integral problem: 1/((x^2)(sqrt(4-x^2)))

Palmira

∫ dx/ x2 (√4-x2)

Homework Equations

x=2sinθ
dx=2cosθdx
sin2 θ= (1 + cos (2θ))/2

The Attempt at a Solution

First attempt:
Plug in the trig sub and get:

∫2cosθdθ/4sin2θ2cosθ
The two cos θ cancel leaving

∫ dθ/ (4sin2θ)

Now I replace the sin2θ with (1-cos2θ)/2
and get

1/2 ∫dθ/(1-cos2θ)

I use u sub with the cos2θ
u=2θ
du/2=dθ

∫cosudu= sin2θ/2

I plug it back into the original equation

1/2 [ 2/θ-sin2θ]= 1/(θ-2sinθcosθ)

I retrieve the x from the original equation to get

4/ arcsin(x/2) -(x)(√(4-x2))

I may be doing something completely wrong, but I can't figure out where. Any hint on where to find the mistake? I keep practicing trig sub problems and I can not get any right. It might have to do with the derivation of the equations, but I am not sure.

Last edited:

Homework Helper

∫ dx/ x2 (√4-x2)

Homework Equations

x=2sinθ
dx=2cosθdx
sin2 θ= (1 + cos (2θ))/2

The Attempt at a Solution

First attempt:
Plug in the trig sub and get:

∫2cosθdθ/4sin2θ2cosθ
The two cos θ cancel leaving

∫ dθ/ (4sin2θ)

Now I replace the sin2θ with (1-cos2θ)/2
and get

1/2 ∫dθ/(1-cos2θ)

I use u sub with the cos2θ
u=2θ
du/2=dθ

∫cosudu= sin2θ/2

I plug it back into the original equation

1/2 [ 2/θ-sin2θ]= 1/(θ-2sinθcosθ)

I retrieve the x from the original equation to get

4/ arcsin(x/2) -(x)(√(4-x2))

I may be doing something completely wrong, but I can't figure out where. Any hint on where to find the mistake? I keep practicing trig sub problems and I can not get any right. It might have to do with the derivation of the equations, but I am not sure.

When you say "I use u sub with the cos2θ" things start going badly wrong. I'm not even sure what you are doing after that. I would go back to when you have 1/(4sin^2(θ)). That's the same as csc^2(θ)/4. There's an easy integral for csc^2(θ).

Palmira
Well I do not feel very smart.

∫csc2xdx/ 4= -cot x/4 +c

go back to the triangle and it is -(√4-x2/ 4x) +c

Thank-you so much!