anonymous
  • anonymous
integrate dx/(4+x^2)^2 using trig substituition. im confused i thought it had to be a square root to be used no squared can some explain how toset these up?
Mathematics
katieb
  • katieb
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zepdrix
  • zepdrix
\[\large\rm \int\limits \frac{dx}{(4+x^2)^2}\]Ummm trig sub? Yah that'll be fun :)
zepdrix
  • zepdrix
If you factor a 4 out of each term in the denominator you get\[\large\rm \int\limits\limits \frac{dx}{(4+x^2)^2}=\frac{1}{4^2}\int\limits\frac{dx}{\left(1+\frac{1}{4}x^2\right)}\]Let's bring the 1/4 into the square with the x,\[\large\rm =\frac{1}{16}\int\limits\limits\frac{dx}{\left(1+\left[\frac{x}{2}\right]^2\right)}\]Now we have that thing in the form: \(\large\rm 1+stuff^2\) So we can make the substitution: \(\large\rm stuff=\tan\theta\)
zepdrix
  • zepdrix
\[\large\rm \frac{x}{2}=\tan\theta\]From there, find your \(\large\rm d\theta\). Plug in all the stuff. Then we'll use some cool trig rules to simplify everything down, and hopefully, be able to integrate easily.

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zepdrix
  • zepdrix
From there, find your \(\large\rm dx\)* is what I meant to say. Any trouble finding it? :o
zepdrix
  • zepdrix
Or were my steps confusing?
anonymous
  • anonymous
Im sorry I just did not understand the find dtheta part. I converted (1+tan0) to sec^20 can i take the derivative of that to dnd d0?
zepdrix
  • zepdrix
\[\large\rm =\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\left[\color{orangered}{\frac{x}{2}}\right]^2\right)^{2}}=\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\color{orangered}{\tan^2\theta}\right)^{2}}=\frac{1}{16}\int\limits\frac{dx}{\sec^4\theta}\]So you've got that much figured out so far? To find dx, you need to go back to your original substitution. \(\large\rm \frac{x}{2}=\tan\theta\) Take derivative of that thing.
anonymous
  • anonymous
so it would turn do x=2tantheta and the derivative is sec^2theta do it would be dtheta over sec^2theta
zepdrix
  • zepdrix
Mmm are you missing a 2 in your dx maybe? Should look like this.\[\large\rm x=2\tan theat\qquad\to\qquad \color{royalblue}{dx=2\sec^2\theta~d\theta}\] \[\large\rm \frac{1}{16}\int\limits\limits\frac{\color{royalblue}{dx}}{\sec^4\theta}=\frac{1}{16}\int\limits\limits\frac{\color{royalblue}{2\sec^2\theta~d\theta}}{\sec^4\theta}\]
zepdrix
  • zepdrix
Ah I made a typo in my x :) lol
zepdrix
  • zepdrix
\[\large\rm =\frac{2}{16}\int\limits\frac{d \theta}{\sec^2\theta}\]So something like this? Ya you have the right idea.
zepdrix
  • zepdrix
Recall that \[\large\rm \cos\theta=\frac{1}{\sec\theta}\]
anonymous
  • anonymous
yeah i have that! and i would just u sub right?!
zepdrix
  • zepdrix
\[\large\rm =\frac{1}{8}\int\limits \cos^2\theta~d \theta\] Hmm no. Looks like you'll have to use your `Half-Angle Formula` at this point. :)
anonymous
  • anonymous
i got 1/8 * integral of cos^2theta
anonymous
  • anonymous
\[=\frac{ 1 }{ 16 }\int\limits \left( 2\cos ^2 \theta \right)d \theta=\frac{ 1 }{ 16 }\int\limits \left( 1+\cos 2\theta \right) d \theta =?\]
anonymous
  • anonymous
its going to be theta -sin2theta/2 but that would make the denominator for sin 32 and it is wrong @surhithayer
zepdrix
  • zepdrix
No, that sounds fine :) You just need your final answer in terms of \(\large\rm x\), not \(\large\rm \theta\). So you need to undo your substitution, perhaps with the use of a triangle.
anonymous
  • anonymous
\[\int\limits \cos 2\theta d \theta =\frac{ \sin 2\theta }{ 2}\]
anonymous
  • anonymous
\[=\frac{ 1 }{ 2 }*\frac{ 2 \tan \theta }{ 1-\tan ^2 \theta }=\frac{ \tan \theta }{ 1-\tan ^2 \theta }\]

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