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Annoying Integral\[\int\frac{dx}{(x^2+a^2)(x^2+b^2)^{1/2}}\]

Mathematics
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What's dx doing in the numerator?
oh nvm
substituting x = c + t, i think we could use Euler substitution ... but not sure if that will work http://en.wikipedia.org/wiki/Euler_substitution

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Here's a wall of math. I seriously hope I didn't make any large mistakes. You'll see that I skipped a lot of work, which I did on paper. I hope I made clear what substitutions I made. I also didn't find the actual result; I just reduced and subbed until I got a manageable integral. \[\int\frac{dx}{(x^2+a^2)(x^2+b^2)^\frac{1}{2}}\] \[x=b\tan t\\ dx=b\sec^2t\;dt\] \[\frac{b}{|b|}\int\frac{\sec t}{b^2\tan^2t+a^2}\;dt\\ \frac{b}{|b|}\int\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\;dt\] \[u=\tan\left(\frac{t}{2}\right)\\ du=\frac{1}{2}\sec^2\left(\frac{t}{2}\right)\\ 2\cos^2\left(\frac{t}{2}\right)du=dt\\ \frac{2}{1+u^2}du=dt\\ \text{We also have}\\ \sin^2t=\frac{4u^2}{(1+u^2)^2}\\ \cos^2t=\frac{(1-u)^2}{(1+u^2)^2}\] \[\frac{b}{|b|}\int\frac{\frac{1-u}{1+u^2}}{\frac{4b^2u^2}{(1+u^2)^2}+\frac{a^2(1-u)^2}{(1+u^2)^2}}\cdot\frac{2}{1+u^2}du\\ \frac{2b}{|b|}\int\frac{\frac{1-u}{(1+u^2)^2}}{\frac{4b^2u^2+a^2(1-u)^2}{(1+u^2)^2}}du\\ \frac{2b}{|b|}\int\frac{1-u}{4b^2u^2+a^2(1-u)^2}du\] \[s=1-u\\ -ds=du\] \[-\frac{2b}{|b|}\int\frac{s}{4b^2(1-s)^2+a^2s^2}ds\\ \small\text{(Completing the square, we get)}\\ -\frac{2b}{|b|}\int\frac{s}{(a^2+4b^2)\left(s-\frac{4b^2}{a^2+4b^2}\right)^2+\left(4b-\frac{16b^4}{a^2+4b^2}\right)}ds\] \[\left(s-\frac{4b^2}{a^2+4b^2}\right)=\frac{ \sqrt{4b-\frac{16b^4}{a^2+4b^2}} }{\sqrt{a^2+4b^2}}\tan r\\ \text{Denote }c=\frac{ \sqrt{4b-\frac{16b^4}{a^2+4b^2}} }{\sqrt{a^2+4b^2}}\\ ds=c\sec^2r\;dr\] \[-\frac{2b}{|b|}\int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{(a^2+4b^2)(c\tan r)^2+\left(4b-\frac{16b^4}{a^2+4b^2}\right)}(c\sec^2r\;dr)\\ -\frac{2b}{|b|}\int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{\left(4b-\frac{16b^4}{a^2+4b^2}\right)\tan^2r+\left(4b-\frac{16b^4}{a^2+4b^2}\right)}(c\sec^2r\;dr)\\ -\frac{2b}{|b|}\cdot\frac{c}{\left(4b-\frac{16b^4}{a^2+4b^2}\right)} \int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{\sec^2r}(\sec^2r\;dr)\\ -\frac{2b}{|b|}\cdot\frac{c}{\left(4b-\frac{16b^4}{a^2+4b^2}\right)} \int\left(c\tan r+\frac{4b^2}{a^2+4b^2}\right) \;dr\]
I can't see how that will priduce the answer that I am given, which is\[\frac1{a\sqrt{b^2-a^2}}\tan^{-1}\sqrt{\frac{b^2-a^2}{a^2(x^2+b^2)}},~~b^2>a^2\]
Wow ... great effort \[ \int\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\;dt = \int\frac{\cos t}{b^2\sin^2t+a^2(1 - \sin^2t )}\;dt \] let sin t= u, ... this should work
but yes, great effort :) It may be right if I can simplify that stuff, but that is scaring me right now :P
that lower term simplifies down to 1/(x^2+a^2) giving arctan.
@experimentX, I can't believe I haven't considered that sub. I really took the long route there.
yeah ... you choose Weierstrass subs ... it gives solution, but this isn't much better than Euler subs.

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