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

  • Thread starter Thread starter Palmira
  • Start date Start date
  • Tags Tags
    Integral
Palmira
Messages
2
Reaction score
0

Homework Statement



∫ 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:
Physics news on Phys.org
Palmira said:

Homework Statement



∫ 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).
 
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!
 
Palmira said:
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!

You're welcome and well done. You started ok. You just missed where to get off the substitution train.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top