Integration Help: Solving x^2/sqrt(16-x^2) Integral

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In summary, the conversation revolves around a problem with trig substitution in integration. The integral in question is (x^2/sqrt(16-x^2)) dx and the correct answer is 8arcsin(x/4) - 0.5x*sqrt(16-x^2). The conversation involves discussing the steps taken to solve the problem, including substituting x=4sinu and dx=4cosu du. The mistake made was in the denominator, which should have been 4cosu instead of 4cos^2u. There is also a brief discussion about not giving out full solutions and repeating information in a more obvious manner.
  • #1
VantagePoint72
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Hello,
I'm currently learning integration by trig substitution. The problem is:
integral ( x^2/sqrt(16-x^2) ) dx.
I seem to be having some difficulty...this is what I have:
let x=4sinu
dx = 4cosu du

Thus, we have:

integral 16sin^2(u)*4cosu du/(4cos^2(u)) (the last part is from the trig identity sin^2x + cos^2x = 1)

But now what? I simplied that as much as I could and then used integration by parts, but I ended up with the wrong answer. The correct answer is 8arcsin(x/4) - 0.5x*sqrt(16-x^2). Have I made a mistake in what I've done so far? If not, where do I proceed?

Thanks,
LOS
 
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  • #2
[tex]\sqrt{1-\sin^2u}=|\cos u|[/tex].
 
  • #3
Check your denominator again. Should the cosine term really be squared? Shouldn't be any need for integration by parts after that.
 
  • #4
[tex]\int \frac{x^2}{\sqrt{16-x^2}}dx[/tex]

[tex]x=4sin u, dx=4cos u du[/tex]

[tex]\int\frac{16sin^2u}{(4cosu)}4cosu du[/tex]

[tex]16\int sin^2udu[/tex]

[tex]16(\frac{u}{2}-\frac{sinu*cosu}{2})[/tex]

[tex]8u-8sinu*cosu[/tex]

[tex]8sin^{-1}(\frac{x}{4})-2x\sqrt{1-(\frac{x}{4})^2}[/tex]

[tex]8sin^{-1}(\frac{x}{4})-2x\frac{\sqrt{16-x^2}}{4}[/tex]

[tex]8sin^{-1}(\frac{x}{4})-\frac{x}{2}\sqrt{16-x^2}[/tex]

Your mistake was that your denominator should have been 4cosu, not [tex]4cos^2u[/tex]
 
  • #5
Your new so I won't say much, but please 1) Do not give out a full solution, it never helps the original poster learn anything and 2) Repeat what another member says just in a more blatantly obvious manner.
 
  • #6
Sorry and thanks
 
  • #7
I feel bad for being so harsh now :( Sorry lol.
 
  • #8
Gib Z said:
I feel bad for being so harsh now :( Sorry lol.
I don't.
I'll flay him him next time.

Flebbyman: I know your adress..:devil:
 
  • #9
Gib Z said:
I feel bad for being so harsh now :( Sorry lol.

Don't sweat it

arildno said:
I know your adress..:devil:

:smile:
 

Related to Integration Help: Solving x^2/sqrt(16-x^2) Integral

1. What is the formula for solving the integral of x^2/sqrt(16-x^2)?

The formula for solving this integral is ∫x^2/√(16-x^2) dx = -√(16-x^2) + C.

2. How do I approach solving this type of integral?

To solve this integral, you can use the substitution method. Let u = √(16-x^2) and solve for du. Then, substitute u and du into the original integral and solve.

3. Can I use any other method to solve this integral?

Yes, you can also use trigonometric substitution by letting x = 4sinθ or x = 4cosθ. This method may be useful if the integral contains a square root of a quadratic expression.

4. What is the range of values for x in this integral?

The range of values for x is -4 ≤ x ≤ 4, as the denominator cannot be equal to 0.

5. Is there a way to check if I have solved the integral correctly?

Yes, you can check your solution by taking the derivative of your answer and seeing if it matches the original integrand. You can also use online integral calculators to verify your solution.

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