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anonymous

  • one year ago

I have a test tmrw and I have NO CLUE how to solve these integrals! example: integrate dx/x(sqrt(x^2+16)) and integrate x^3(sqrt(9-x^2))dx can someone give me some help?

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  1. zepdrix
    • one year ago
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    Oooo trig stuff :O fun!

  2. Jhannybean
    • one year ago
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    \[\int \frac{dx}{x(\sqrt{x^2+16})}\] I believe this follows one of the basic integral forms of \(\sqrt{a^2+u^2}\)\[\sf \int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a}\ln \left|\frac{\sqrt{a^2+u^2}}{u}\right|+C\]

  3. anonymous
    • one year ago
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    Am I supposed to be seeing an equation? sorry. on the second one I got

  4. anonymous
    • one year ago
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    integral u(sqrt(9-u))du where u=x^2 and 1/2du=xdx

  5. zepdrix
    • one year ago
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    Hmm I don't think that u-sub is going to be very effective. If you insist on doing a u-sub, you'll want to do \(\large\rm u=9-x^2\) But trig sub would be a lot better :O

  6. Jhannybean
    • one year ago
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    hmm.. thats part of it. \[\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int\color{blue}{ x}\cdot \color{red}{x^2}\sqrt{9-\color{red}{x^2}}\color{blue}{dx}\]\[\color{red}{u=x^2}~,~ \color{blue}{du=2xdx \implies xdx =\dfrac{1}{2}du} \]

  7. anonymous
    • one year ago
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    zepdrix (sorry Jhannybean) can you give me a push as to what a should be for the second problem?

  8. zepdrix
    • one year ago
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    Bahhh I take back what I said :) The u-sub I mentioned should work out really nicely! \(\large\rm u=9-x^2\)

  9. Jhannybean
    • one year ago
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    Ah ha... I just read up on the trig subs. \[\sqrt{9-x^2} \qquad \implies\qquad \sqrt{a^2-x^2} ~,~ x=a\sin(\theta)\]This!!! Now I recall.

  10. zepdrix
    • one year ago
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    You do that because you really would like to get rid of the subtraction under the root. \(\large\rm du=-2x~dx\qquad\to\qquad -\frac{1}{2}du=x~dx\) But also:\[\large\rm u=9-x^2\qquad\to\qquad x^2=u+9\] And then sub all your pieces in! :)\[\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)\]

  11. anonymous
    • one year ago
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    Oh boy... Can anybody teach me how to set up the "triangle" to know to use cos, sin or tan?

  12. Jhannybean
    • one year ago
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    Therefore \[\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int(3\sin(\theta))^3\sqrt{9-(3\sin(\theta))^2 }\cdot 3\cos(\theta)d\theta\] Something like that?...

  13. Jhannybean
    • one year ago
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    Where \(x=3\sin(\theta) ~,\qquad ~ dx = 3\cos(\theta)d\theta\)

  14. anonymous
    • one year ago
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    ok... I only know how to do a trig subs with fractions where dx is on top...

  15. Jhannybean
    • one year ago
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    Wow,yeah I agree @zepdrix , this is going to be really long x_x haha crap.

  16. zepdrix
    • one year ago
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    I'll group the steps together that I posted earlier, maybe it will help :o

  17. zepdrix
    • one year ago
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    \[\large\rm \int\limits\limits x^3\sqrt{9-x^2}~dx=\int\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})\]Making the substitution:\[\large\rm \color{#CC0033 }{u=9-x^2}\]Adding x^2 to each side, subtracting u from each side gives us,\[\large\rm \color{#F35633}{x^2=9-u}\]Differentiating the red one gives us,\[\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=dx}\]\[\large\rm \int\limits\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})=\int\limits\limits \color{#F35633}{(9-u)}\sqrt{\color{#CC0033}{u}}~\left(\color{green}{-\frac{1}{2}du}\right)\]

  18. zepdrix
    • one year ago
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    From there it's not too bad, ya? :) Rewrite your square root as a 1/2 power,\[\large\rm =-\frac{1}{2}(9-u)u^{1/2}~du\]Distribute the u^(1/2), and then just apply your power rule to each term.

  19. zepdrix
    • one year ago
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    Doh :d typo on the green step, my bad\[\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=x~dx}\]

  20. anonymous
    • one year ago
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    HEY! What if I set u^2=9-x^2? because then the sqrt is resolved to just u, and I'm left with integrate: (9-u^2)(-u^2)du?

  21. zepdrix
    • one year ago
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    That's a really strange way to approach it :D But I think it works! Let's see...\[\large\rm u^2=9-x^2,\qquad\quad x^2=u^2+9,\qquad\quad u=\sqrt{9-x^2}\]Differentiating the first part,\[\large\rm 2u~du=-2x~dx\qquad\to\qquad -u~du=x~dx\]Subbing everything in gives you,\[\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)=\int\limits (u^2+9)\cdot u\cdot (-u~du)\]I think maybe your sign is off in the middle, looks good besides that though.

  22. zepdrix
    • one year ago
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    Naw your sign is good :) mine was bad lol

  23. anonymous
    • one year ago
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    I worked it out both waym and they both check with the answer in the book! Any help on the first one? I know it has to do with x=sina or something?

  24. zepdrix
    • one year ago
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    \[\large\rm \int\limits \frac{dx}{x \sqrt{x^2+16}}\]You would really like to get \(\large\rm (stuff)^2+1\) under the root, then you could make the substitution:\[\large\rm stuff=\tan\theta\]And we would get,\[\large\rm (\tan \theta)^2+1=\sec^2\theta\]Trig sub is a clever way to get rid of the addition/subtraction between terms! It allows us to easily take this out of the root.

  25. zepdrix
    • one year ago
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    How do we get +1 though? Hmm we have a +16... Well, if we can make a 16 show up in our "stuff", then we can factor a 16 out of each term, leaving us with 16(stuff^2+1) under the root.

  26. zepdrix
    • one year ago
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    Hopefully that makes some sense :p I know I'm throwing the word stuff around a lot...

  27. zepdrix
    • one year ago
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    So it seems that sine isn't the sub we're looking for. Since we have `addition`, we're going to head over to tangent, since the `square identity` involving tangent has addition in it.

  28. anonymous
    • one year ago
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    Looking around, I found this helpful powerpiont. It gives a good start to anyone looking at this thread when its archived: https://drive.google.com/file/d/0B0lXunoGraQaQ2tfZ0NCbFpUeDQ/edit

  29. anonymous
    • one year ago
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    Speaking of, we need an a(tan)theta = x for this one.

  30. zepdrix
    • one year ago
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    lol

  31. anonymous
    • one year ago
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    give me a second to work this out a bit...

  32. Astrophysics
    • one year ago
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    Did you work it out? I like this question so I'll show a solution you can then compare your answer to.. \[\int\limits \frac{ dx }{ x \sqrt{x^2+16} }\] \[x= \tan \theta \implies dx = \sec^2\theta d \theta\] that will be our substitution, so our new integral will look like \[\int\limits \frac{ 4\sec^2\theta d \theta }{ \tan \theta \sqrt{(4\tan \theta)^2+16} }\] now notice in the denominator we can factor out the 16, \[\int\limits \frac{ 4 \sec^2 \theta d \theta }{ 4\tan \theta \sqrt{16 \tan^2 \theta+16} } \implies \int\limits \frac{ \sec^2 \theta d \theta }{ 4 \tan \theta \sqrt{16(\tan^2 \theta+1)} }\] now we can use the following identity \[\tan^2 \theta +1 = \sec^2 \theta\] so we get \[\int\limits \frac{ \sec^2 \theta d \theta }{ \tan \theta \sqrt{16 \sec^2 \theta} } \implies \frac{ 1 }{ 4 }\int\limits \frac{ \sec^2 \theta }{ \tan \theta \sec \theta }d \theta\]\[\frac{ 1 }{ 4 } \int\limits \frac{ \sec \theta d \theta }{ \tan \theta } d \theta \implies \frac{ 1 }{ 4 } \int\limits \csc \theta d \theta\] and remember the integral of csc theta is \[\frac{ 1 }{ 4 } \int\limits \csc \theta d \theta = \frac{ 1 }{ 4 } \ln|\csc \theta-\cot \theta|+C\] but we're not exactly done yet as we made a substitution earlier. |dw:1443826146119:dw| looking at the trig ratios we finally get \[\frac{ 1 }{ 4 }(\ln|x|-\ln|\sqrt{x^2+16}+4|)+C\]

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