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How to integrate the last two terms in that integral

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  • #1
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I'm having trouble with integrating this. [tex]\int_1^8 \sqrt{9x^{4}+\frac{1}{2}+\frac{1}{144x^{4}}} dx[/tex]

I can get the problem down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex] , but I don't know how to integrate the last two terms in that integral. Am I doing something wrong?
 

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  • #2
VietDao29
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I'm having trouble with integrating this. [tex]\int_1^8 \sqrt{9x^{4}+\frac{1}{2}+\frac{1}{144x^{4}}} dx[/tex]

I can get the problem down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex] , but I don't know how to integrate the last two terms in that integral. Am I doing something wrong?
How did you get it down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex]? It doesn't look correct to me. Can you show us the steps so that we can check them for you?

Hint: Since there's a big square root, one should first try to Complete the Square.
 
  • #3
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I distributed the (1/2) power among everything under the radical. Is that incorrect?
 
  • #4
SammyS
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I distributed the (1/2) power among everything under the radical. Is that incorrect?
Is [tex]\sqrt{1+1+1}=\sqrt{1}+\sqrt{1}+\sqrt{1}\ ?[/tex]

I so , then [tex]\sqrt{3}=3\,.[/tex]
 
  • #5
eumyang
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I distributed the (1/2) power among everything under the radical. Is that incorrect?
Yes, that's completely wrong. You can't do that with a sum inside a square root.

Hint: Since there's a big square root, one should first try to Complete the Square.
I would instead find a common denominator and add the three fractions.
 
  • #6
VietDao29
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I distributed the (1/2) power among everything under the radical. Is that incorrect?
You seem not to know how to Complete the Square.

Roughly speaking, Complete the Square is the manipulation you do to change your expression into something like: (ax + b)2 + c2. Where a, b, and c are all constant.

Say, your expression is:
[tex]4x ^ 2 + 8x + 8[/tex]

Here's how you would do it. You'll try to split your expression into a square. Note that:
(a + b)2 = a2 + 2ab + b2

[tex]4x ^ 2 + 8x + 8 = (2x) ^ 2 + 8x + 8 = (2x) ^ 2 + 2.(2x).2 + 8 = \left[(2x) ^ 2 + 2.(2x).2 + 2 ^ 2 \right] + 4 = (2x + 2) ^ 2 + 4[/tex].

Another example is:
[tex]x ^ 2 + 5 + \frac{1}{4x ^ 2} = x ^ 2 + \left( \frac{1}{2x} \right) ^ 2 + 5 = x ^ 2 + 2.x.\frac{1}{2x} + \left( \frac{1}{2x} \right) ^ 2 + 4 = \left( x + \frac{1}{2x} \right) ^ 2 + 4[/tex].

Let's see if you can Complete the Square for:
[tex]9x^{4}+\frac{1}{2}+\frac{1}{144x^{4}}[/tex] :)
 
  • #7
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I do not know how to complete the square since x is in the denominator. I can get it to [tex](3x^{2})^{2}+(\frac{1}{12x^{2}})^{2}= -\frac{1}{2}[/tex]
 
  • #8
VietDao29
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I do not know how to complete the square since x is in the denominator. I can get it to [tex](3x^{2})^{2}+(\frac{1}{12x^{2}})^{2}= -\frac{1}{2}[/tex]
The identity we should use here is:
(a + b)2 = a2 + 2ab + b2

In your problem:
3x2 is a
1/(12x2) is b.

So, what should you add to make it becomes (a + b)2?

Don't make it become an equation like this: [tex](3x^{2})^{2}+(\frac{1}{12x^{2}})^{2}= -\frac{1}{2}[/tex]

You can look again at my 2 examples to have some ideas.
 
  • #9
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So 2ab is 1/2.
Then if we have the square root of (a+b)^2, that equals (a+b), and b still has x in the denominator, so how do I integrate that?
 
  • #10
eumyang
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Rewrite
[tex]\frac{1}{12x^2}[/tex]
as an expression with a negative exponent, and then use the power rule.
 
  • #11
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Wow. I'm really off of my game tonight. I finally got it worked out. Thank you everyone for your help. I truly appreciate it.
 

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