# How can these two be equivalent (wolfram-alpha incorrect) !?

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!)

• 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? – Alex Becker Dec 11 '13 at 2:33
• Also, what is b? – Alex Becker Dec 11 '13 at 2:33
• Facepalm And $b$ was a typo. – user1833028 Dec 11 '13 at 2:38
• When you're dealing with trigonometric functions, it's very easy to forget the basic underlying structure: their relationships on a right triangle. – Lost Dec 11 '13 at 2:58
• Welcome to trigonometry. – Michael Hardy Dec 11 '13 at 4:03

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 $1/a$.
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 $|a| < |x|$.