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So wolfram-alpha reads The integral of $$\int \frac{1}{\sqrt{a^2-x^2}}dx=\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$$ but that $$\int\frac{1}{\sqrt{a^2-x^2}}dx \;\mathrm{where}\; a=5 \;\mathrm{is}\; \sin^{-1}{(\frac{x}{5})}$$

I have no idea why the integration varies based on a cleanly pluggable value of a. Could someone be so kinda as to explain what just happened?

(I ran $\mathrm{is }\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})\;\mathrm{equal}\;\mathrm{to}\sin^{-1}(\frac{x}{a})$, and the result was false!)

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  • $\begingroup$ Why didn't you check whether $$\tan^{-1}\left(\frac{x}{5^2-x^2}\right)=\sin^{-1}\left(\frac{x}{5}\right)$$ which is the relevant question? $\endgroup$ – Alex Becker Dec 11 '13 at 2:33
  • $\begingroup$ Also, what is b? $\endgroup$ – Alex Becker Dec 11 '13 at 2:33
  • $\begingroup$ Facepalm And $b$ was a typo. $\endgroup$ – user1833028 Dec 11 '13 at 2:38
  • $\begingroup$ When you're dealing with trigonometric functions, it's very easy to forget the basic underlying structure: their relationships on a right triangle. $\endgroup$ – Lost Dec 11 '13 at 2:58
  • $\begingroup$ Welcome to trigonometry. $\endgroup$ – Michael Hardy Dec 11 '13 at 4:03
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If $$\theta = \tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$$ you have $$ \tan(\theta) = \frac{x}{\sqrt{a^2-x^2}}.$$ Draw the triangle. opposite $= x$, adjacent $=\sqrt{a^2 - x^2}$, and so hypotenuse $=a$. Now compute the trig ratios. The simplest is $\sin(\theta) = {x\over a}.$ I think you are missing a factor of $1/a$.

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These results are of course the same. To see this, plug in: $$x = a\sin \theta$$ And remember that $1 - \sin^2 \theta = \cos^2 \theta$.

The reason wolfram doesn't think so probably has to do with the domain, since the functions are no longer the same if $|a| < |x|$.

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