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Mathematics 15 Online
OpenStudy (anonymous):

Help with an indefinite integral: int frac{ dx }{ x(x+4)(sqrt{4-x^2)} } So, i've tried trig substitution, but i get stuck with some int (dx/(sinx+2). I think is possible to get rid of the radical and do somethig like partial fractions. Any idea?

OpenStudy (anonymous):

\[\int\limits \frac{ dx }{ x(x+4)(\sqrt{4-x^2)} }\]

OpenStudy (xapproachesinfinity):

any ideas yet?

OpenStudy (xapproachesinfinity):

trig sub? can you think of any?

OpenStudy (anonymous):

x=2sin(u)

OpenStudy (xapproachesinfinity):

oh you did trig sub you got \[1/4 \int \tan u\frac{du}{\cos u(\sin u+2)}\]

OpenStudy (xapproachesinfinity):

cotu not tanu *

OpenStudy (xapproachesinfinity):

\[1/4\int \frac{du}{\sin u(\sin u+2)}\]

OpenStudy (xapproachesinfinity):

is that what you had?

OpenStudy (rational):

you may try `t = tan(u/2)` for evaluating the `integral 1/(sin(u)+2)`

OpenStudy (anonymous):

\[1/4 \int\limits du/2\sin(u) - 1/8 \int\limits du/(\sin(u)+2)\]

OpenStudy (anonymous):

The trigo sub sounded like a good idea. Setting \(x=2\sin u\) to get the differential \(dx=2\cos u\,du\), you have \[\begin{align*}\int\frac{dx}{x(x+4)\sqrt{4-x^2}}&=\int\frac{2\cos u}{2\sin u(2\sin u+4)\sqrt{4-4\sin^2u}}\,du\\\\ &=\frac{1}{4}\int\frac{du}{\sin u(\sin u+2)}\\\\ &=\frac{1}{4}\int\left(\frac{f(u)(\sin u+2)}{\sin u(\sin u+2)}+\frac{g(u)\sin u}{\sin u(\sin u+2)}\right)\,du \end{align*}\] Let's try finding \(f(u)\) and \(g(u)\) such that the numerator is equivalent to \(1\). If we make \(f(u)=\dfrac{1}{2}\) and \(g(u)=-\dfrac{1}{2}\), that should do the trick. \[\frac{1}{2}\sin u+1-\frac{1}{2}\sin u=1\] From there, rational's suggestion regarding the half-angle tangent substitution works wonders.

OpenStudy (anonymous):

...which you've apparently figured out on your own :P

OpenStudy (xapproachesinfinity):

try rational's sub

OpenStudy (anonymous):

hmm... well, how can i express sin(u) in terms of t?

OpenStudy (rational):

\(t=\tan(u/2)\) use double angle identity of \(\sin(u)\) in terms of \(\tan(u/2)\) : \[\sin(u) = \frac{2\tan(u/2)}{1+\tan^2(u/2)}=\frac{2t}{1+t^2}\]

OpenStudy (xapproachesinfinity):

u will have to use long division with this one after this part

OpenStudy (rational):

i think we end up with 1/(t^2+t+1) factoring the denominator using quadratic formula, then partial fractions... really a very long path

OpenStudy (anonymous):

it should be an easier way. Anyway. Thanks a lot.

OpenStudy (rational):

np:)

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