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genius12
 one year ago
Best ResponseYou've already chosen the best response.0What's dx doing in the numerator?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.2substituting x = c + t, i think we could use Euler substitution ... but not sure if that will work http://en.wikipedia.org/wiki/Euler_substitution

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.3Here'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. \[\int\frac{dx}{(x^2+a^2)(x^2+b^2)^\frac{1}{2}}\] \[x=b\tan t\\ dx=b\sec^2t\;dt\] \[\frac{b}{b}\int\frac{\sec t}{b^2\tan^2t+a^2}\;dt\\ \frac{b}{b}\int\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\;dt\] \[u=\tan\left(\frac{t}{2}\right)\\ du=\frac{1}{2}\sec^2\left(\frac{t}{2}\right)\\ 2\cos^2\left(\frac{t}{2}\right)du=dt\\ \frac{2}{1+u^2}du=dt\\ \text{We also have}\\ \sin^2t=\frac{4u^2}{(1+u^2)^2}\\ \cos^2t=\frac{(1u)^2}{(1+u^2)^2}\] \[\frac{b}{b}\int\frac{\frac{1u}{1+u^2}}{\frac{4b^2u^2}{(1+u^2)^2}+\frac{a^2(1u)^2}{(1+u^2)^2}}\cdot\frac{2}{1+u^2}du\\ \frac{2b}{b}\int\frac{\frac{1u}{(1+u^2)^2}}{\frac{4b^2u^2+a^2(1u)^2}{(1+u^2)^2}}du\\ \frac{2b}{b}\int\frac{1u}{4b^2u^2+a^2(1u)^2}du\] \[s=1u\\ ds=du\] \[\frac{2b}{b}\int\frac{s}{4b^2(1s)^2+a^2s^2}ds\\ \small\text{(Completing the square, we get)}\\ \frac{2b}{b}\int\frac{s}{(a^2+4b^2)\left(s\frac{4b^2}{a^2+4b^2}\right)^2+\left(4b\frac{16b^4}{a^2+4b^2}\right)}ds\] \[\left(s\frac{4b^2}{a^2+4b^2}\right)=\frac{ \sqrt{4b\frac{16b^4}{a^2+4b^2}} }{\sqrt{a^2+4b^2}}\tan r\\ \text{Denote }c=\frac{ \sqrt{4b\frac{16b^4}{a^2+4b^2}} }{\sqrt{a^2+4b^2}}\\ ds=c\sec^2r\;dr\] \[\frac{2b}{b}\int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{(a^2+4b^2)(c\tan r)^2+\left(4b\frac{16b^4}{a^2+4b^2}\right)}(c\sec^2r\;dr)\\ \frac{2b}{b}\int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{\left(4b\frac{16b^4}{a^2+4b^2}\right)\tan^2r+\left(4b\frac{16b^4}{a^2+4b^2}\right)}(c\sec^2r\;dr)\\ \frac{2b}{b}\cdot\frac{c}{\left(4b\frac{16b^4}{a^2+4b^2}\right)} \int\frac{c\tan r+\frac{4b^2}{a^2+4b^2} }{\sec^2r}(\sec^2r\;dr)\\ \frac{2b}{b}\cdot\frac{c}{\left(4b\frac{16b^4}{a^2+4b^2}\right)} \int\left(c\tan r+\frac{4b^2}{a^2+4b^2}\right) \;dr\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.0I can't see how that will priduce the answer that I am given, which is\[\frac1{a\sqrt{b^2a^2}}\tan^{1}\sqrt{\frac{b^2a^2}{a^2(x^2+b^2)}},~~b^2>a^2\]

experimentX
 one year ago
Best ResponseYou've already chosen the best response.2Wow ... 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

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.0but yes, great effort :) It may be right if I can simplify that stuff, but that is scaring me right now :P

experimentX
 one year ago
Best ResponseYou've already chosen the best response.2that lower term simplifies down to 1/(x^2+a^2) giving arctan.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.3@experimentX, I can't believe I haven't considered that sub. I really took the long route there.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.2yeah ... you choose Weierstrass subs ... it gives solution, but this isn't much better than Euler subs.
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