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TuringTest
Group Title
Annoying Integral\[\int\frac{dx}{(x^2+a^2)(x^2+b^2)^{1/2}}\]
 one year ago
 one year ago
TuringTest Group Title
Annoying Integral\[\int\frac{dx}{(x^2+a^2)(x^2+b^2)^{1/2}}\]
 one year ago
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

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

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

SithsAndGiggles Group TitleBest 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.
 one year ago

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