Integral help needed

  • Thread starter Melawrghk
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  • #1
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Homework Statement


[tex]\int \frac{26 dx}{{(169x^2+1)}^2}[/tex] <= the whole denominator is supposed to be squared...

The Attempt at a Solution


So I converted the thing in the denominator so that it has a square root:
[tex]\int \frac{26 dx}{{\sqrt{169x^2+1}}^4}[/tex]

Looking at the denominator, I realized I should do an inverse substitution:
13x=tan(t)
dx=sec2(t)*dt/13

I subbed that in to the equation before and got:
[tex]\int \frac{26 * sec^2(t) * dt}{13*sec^4(t)}[/tex]

Simplifying which, I get:
2[tex]\int cos^2(t) dt[/tex]

Then I tried doing integration by parts, but I got nowhere - I kept getting cos^2 again... Please help me, this question frustrates me. Thanks in advance!

(Sidenote: I finally got my formulas all pretty, yay!)
 

Answers and Replies

  • #2
Dick
Science Advisor
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There's not really much profit into converting x^2 into sqrt(x)^4, is there? The rest of the general approach looks fine. You get sec^2/sec^4. So sure, cos^2(x). You just want to use a double angle formula cos(x)^2=(1+cos(2x))/2.
 
  • #3
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Thanks! I know there isn't a point with the root, I'm just more used to seeing it like that. And thanks for the double angle formula. I always forget they exist... I guess it's time I memorize them :)
 

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