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

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SUMMARY

The integral problem discussed involves the evaluation of the integral ∫ dx/(x²√(4-x²)). The user initially attempted a trigonometric substitution with x=2sinθ, leading to an incorrect path after substituting and simplifying. The correct approach involves recognizing that 1/(4sin²(θ)) can be expressed as (csc²(θ))/4, which has a straightforward integral. The final result is -√(4-x²)/(4x) + C, highlighting the importance of correctly identifying when to stop substitutions in trigonometric integrals.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sin²(θ) and csc²(θ).
  • Proficiency in performing trigonometric substitutions in integrals.
  • Familiarity with integration techniques, including u-substitution and basic integral forms.
  • Knowledge of inverse trigonometric functions and their applications in integration.
NEXT STEPS
  • Study the integral of csc²(θ) to solidify understanding of its application in integration.
  • Practice additional problems involving trigonometric substitution to gain confidence.
  • Review the derivation and application of inverse trigonometric functions in calculus.
  • Explore common pitfalls in integration techniques to avoid errors in future problems.
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of trigonometric substitution in action.

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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.
 
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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.
 

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