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Very difficult integral to solve

  • Thread starter tonit
  • Start date
  • #1
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Homework Statement


hey there guys, today I encountered a very difficult integral to solve, at least for me.
[itex]\int^\sqrt{3}_1 \frac{dx}{\sqrt{(1+x^2)^3}}[/itex]


Homework Equations





The Attempt at a Solution


I've tried to substitute but didn't give results. Any suggestions please :)
 

Answers and Replies

  • #2
Dick
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What substitution did you try? I'd recommend a trig substitution, like x=tan(t).
 
  • #3
55
1
Hmmm, alright I'm trying it now.
 
  • #4
55
1
I'm now stuck at this point:

[itex]\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^3(t)}[/itex]
 
  • #5
Dick
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I'm now stuck at this point:

[itex]\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^3(t)}[/itex]
You'll need to show how you got there. That's not right and I can't guess what you did.
 
  • #6
55
1
x = tan(t)
dx = d(tan[t]) = [itex]\frac{dt}{cos^2(t)}[/itex]
x = 1 => tan(t) = 1 => t = [itex]\frac{\pi}{4}[/itex]
x = [itex]\sqrt{3}[/itex] => tan(t) = [itex]\sqrt{3}[/itex] => t = [itex]\frac{\pi}{3}[/itex]

so the integral becomes
[itex]\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^2(t)\sqrt{(1+tan^2(t))^3}}[/itex]

oh and now I spot my mistake. It will become cos(t)dt, right?
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
618
x = tan(t)
dx = d(tan[t]) = [itex]\frac{dt}{cos^2(t)}[/itex]
x = 1 => tan(t) = 1 => t = [itex]\frac{\pi}{4}[/itex]
x = [itex]\sqrt{3}[/itex] => tan(t) = [itex]\sqrt{3}[/itex] => t = [itex]\frac{\pi}{3}[/itex]

so the integral becomes
[itex]\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^2(t)\sqrt{(1+tan^2(t))^3}}[/itex]

oh and now I spot my mistake. It will become cos(t)dt
Yes it will.
 
  • #8
55
1
Ok, thanks for helping me :D
 

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