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Attempting to calculate $\displaystyle \int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}$, $$\int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}=\int{\frac{1}{\sqrt{(y\tan\theta)^2+y^2}}y\sec^2\theta \mathrm d\theta}=\int{\sec\theta d\theta}=\ln(\sec\theta +\tan\theta)=\ln\left(\sqrt{\left(\frac{x}{y}\right)^2+1}+\frac{x}{y}\right)=\ln\left(\frac{1}{x}\left(\sqrt{x^2+y^2}+x\right)\right),$$ where $x=y\tan\theta$

However Wolfram Integrator somehow returns $$\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$ as the answer. Where did I go wrong? Many thanks.

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  • $\begingroup$ For one, you should have factored out a $1/y$ rather than a $1/x$. The factor of two can be wrapped up in an integration constant. $\endgroup$ – Ron Gordon Oct 5 '13 at 12:21
  • $\begingroup$ The only nitpick I have so far is that the coefficient in front of the natural logarithm at the end of your solution should be $\frac{1}{y}$ and not $1/x$. $\endgroup$ – Clayton Oct 5 '13 at 12:21
  • $\begingroup$ I think the wolfram integrator is wrong see wolframalpha.com/input/… $\endgroup$ – Shobhit Oct 5 '13 at 12:24
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    $\begingroup$ @Shobhit: It isn't wrong, it just chose a (seemingly arbitrary) specific constant of $\ln(2)$. The only thing I dislike is how both have absorbed the $-\ln(y)$ into the constant. $\endgroup$ – Clayton Oct 5 '13 at 12:27
  • $\begingroup$ @Clayton agreed $\endgroup$ – Shobhit Oct 5 '13 at 12:28
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Hint: after you fix the mistake in the last step, differentiate the function $$x\mapsto \ln\left(\frac{1}{y}\left(\sqrt{x^2+y^2}+x\right)\right)-\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$ with respect to $x$. It's easy to do so in your head. Also do not forget that you should consider the absolute value appropriately ($\int \frac 1x\mathrm dx)=\ln (|x|)$) and to add an arbitrary constant when you integrate.

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  • $\begingroup$ It's even easier if you expand the right-hand side to see how much cancellation there is and then differentiate :) $\endgroup$ – Clayton Oct 5 '13 at 12:29
  • $\begingroup$ @Clayton That's exactly what I have in mind. $\endgroup$ – Git Gud Oct 5 '13 at 12:30
  • $\begingroup$ Ah, I apologize for stating what you already knew. I thought you were implying taking the derivative of the function as it looks was very easy, and I personally didn't want to do that. $\endgroup$ – Clayton Oct 5 '13 at 12:32
  • $\begingroup$ @Clayton I'm glad for your comment. If it wasn't clear to you what I meant, it won't be for other people either. $\endgroup$ – Git Gud Oct 5 '13 at 12:32

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