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anonymous

  • one year ago

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?

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  1. zepdrix
    • one year ago
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    \[\large\rm \int\limits \frac{dx}{(4+x^2)^2}\]Ummm trig sub? Yah that'll be fun :)

  2. zepdrix
    • one year ago
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    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: \(\large\rm stuff=\tan\theta\)

  3. zepdrix
    • one year ago
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    \[\large\rm \frac{x}{2}=\tan\theta\]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.

  4. zepdrix
    • one year ago
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    From there, find your \(\large\rm dx\)* is what I meant to say. Any trouble finding it? :o

  5. zepdrix
    • one year ago
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    Or were my steps confusing?

  6. anonymous
    • one year ago
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    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?

  7. zepdrix
    • one year ago
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    \[\large\rm =\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\left[\color{orangered}{\frac{x}{2}}\right]^2\right)^{2}}=\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\color{orangered}{\tan^2\theta}\right)^{2}}=\frac{1}{16}\int\limits\frac{dx}{\sec^4\theta}\]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.

  8. anonymous
    • one year ago
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    so it would turn do x=2tantheta and the derivative is sec^2theta do it would be dtheta over sec^2theta

  9. zepdrix
    • one year ago
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    Mmm are you missing a 2 in your dx maybe? Should look like this.\[\large\rm x=2\tan theat\qquad\to\qquad \color{royalblue}{dx=2\sec^2\theta~d\theta}\] \[\large\rm \frac{1}{16}\int\limits\limits\frac{\color{royalblue}{dx}}{\sec^4\theta}=\frac{1}{16}\int\limits\limits\frac{\color{royalblue}{2\sec^2\theta~d\theta}}{\sec^4\theta}\]

  10. zepdrix
    • one year ago
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    Ah I made a typo in my x :) lol

  11. zepdrix
    • one year ago
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    \[\large\rm =\frac{2}{16}\int\limits\frac{d \theta}{\sec^2\theta}\]So something like this? Ya you have the right idea.

  12. zepdrix
    • one year ago
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    Recall that \[\large\rm \cos\theta=\frac{1}{\sec\theta}\]

  13. anonymous
    • one year ago
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    yeah i have that! and i would just u sub right?!

  14. zepdrix
    • one year ago
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    \[\large\rm =\frac{1}{8}\int\limits \cos^2\theta~d \theta\] Hmm no. Looks like you'll have to use your `Half-Angle Formula` at this point. :)

  15. anonymous
    • one year ago
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    i got 1/8 * integral of cos^2theta

  16. anonymous
    • one year ago
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    \[=\frac{ 1 }{ 16 }\int\limits \left( 2\cos ^2 \theta \right)d \theta=\frac{ 1 }{ 16 }\int\limits \left( 1+\cos 2\theta \right) d \theta =?\]

  17. anonymous
    • one year ago
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    its going to be theta -sin2theta/2 but that would make the denominator for sin 32 and it is wrong @surhithayer

  18. zepdrix
    • one year ago
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    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.

  19. anonymous
    • one year ago
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    \[\int\limits \cos 2\theta d \theta =\frac{ \sin 2\theta }{ 2}\]

  20. anonymous
    • one year ago
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    \[=\frac{ 1 }{ 2 }*\frac{ 2 \tan \theta }{ 1-\tan ^2 \theta }=\frac{ \tan \theta }{ 1-\tan ^2 \theta }\]

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