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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(9x^2))dx
can someone give me some help?
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(9x^2))dx can someone give me some help?

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Oooo trig stuff :O fun!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int \frac{dx}{x(\sqrt{x^2+16})}\] I believe this follows one of the basic integral forms of \(\sqrt{a^2+u^2}\)\[\sf \int \frac{du}{u\sqrt{u^2+a^2}} = \frac{1}{a}\ln \left\frac{\sqrt{a^2+u^2}}{u}\right+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am I supposed to be seeing an equation? sorry. on the second one I got

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0integral u(sqrt(9u))du where u=x^2 and 1/2du=xdx

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm I don't think that usub is going to be very effective. If you insist on doing a usub, you'll want to do \(\large\rm u=9x^2\) But trig sub would be a lot better :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm.. thats part of it. \[\int x^3\sqrt{9x^2}dx \qquad \implies \qquad \int\color{blue}{ x}\cdot \color{red}{x^2}\sqrt{9\color{red}{x^2}}\color{blue}{dx}\]\[\color{red}{u=x^2}~,~ \color{blue}{du=2xdx \implies xdx =\dfrac{1}{2}du} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0zepdrix (sorry Jhannybean) can you give me a push as to what a should be for the second problem?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Bahhh I take back what I said :) The usub I mentioned should work out really nicely! \(\large\rm u=9x^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah ha... I just read up on the trig subs. \[\sqrt{9x^2} \qquad \implies\qquad \sqrt{a^2x^2} ~,~ x=a\sin(\theta)\]This!!! Now I recall.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3You 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:\[\large\rm u=9x^2\qquad\to\qquad x^2=u+9\] And then sub all your pieces in! :)\[\large\rm \int\limits x^2\sqrt{9x^2}~(x~dx)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh boy... Can anybody teach me how to set up the "triangle" to know to use cos, sin or tan?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Therefore \[\int x^3\sqrt{9x^2}dx \qquad \implies \qquad \int(3\sin(\theta))^3\sqrt{9(3\sin(\theta))^2 }\cdot 3\cos(\theta)d\theta\] Something like that?...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where \(x=3\sin(\theta) ~,\qquad ~ dx = 3\cos(\theta)d\theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok... I only know how to do a trig subs with fractions where dx is on top...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow,yeah I agree @zepdrix , this is going to be really long x_x haha crap.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I'll group the steps together that I posted earlier, maybe it will help :o

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3From there it's not too bad, ya? :) Rewrite your square root as a 1/2 power,\[\large\rm =\frac{1}{2}(9u)u^{1/2}~du\]Distribute the u^(1/2), and then just apply your power rule to each term.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Doh :d typo on the green step, my bad\[\large\rm du=2x~dx\qquad\to\qquad \color{green}{\frac{1}{2}du=x~dx}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0HEY! What if I set u^2=9x^2? because then the sqrt is resolved to just u, and I'm left with integrate: (9u^2)(u^2)du?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3That's a really strange way to approach it :D But I think it works! Let's see...\[\large\rm u^2=9x^2,\qquad\quad x^2=u^2+9,\qquad\quad u=\sqrt{9x^2}\]Differentiating the first part,\[\large\rm 2u~du=2x~dx\qquad\to\qquad u~du=x~dx\]Subbing everything in gives you,\[\large\rm \int\limits x^2\sqrt{9x^2}~(x~dx)=\int\limits (u^2+9)\cdot u\cdot (u~du)\]I think maybe your sign is off in the middle, looks good besides that though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Naw your sign is good :) mine was bad lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \int\limits \frac{dx}{x \sqrt{x^2+16}}\]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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3How 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hopefully that makes some sense :p I know I'm throwing the word stuff around a lot...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looking 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Speaking of, we need an a(tan)theta = x for this one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0give me a second to work this out a bit...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Did 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 \[\frac{ 1 }{ 4 }(\lnx\ln\sqrt{x^2+16}+4)+C\]
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