The common method of evaluating the integral below is by using the substitution $x = a \sin u$ which gives: $$\int\frac{dx}{\sqrt{a^2-x^2}}= \arcsin\Bigl(\frac xa\Bigr) + C.$$ But, using this substitution led me down a path that seems to give a different answer (or, at least, a different form). Did I make a mistake, or are the two forms somehow equivalent?
$$x = ai\tan u $$ $$x^2=-a^2\tan^2(u)du$$ $$dx = ai\sec^2(u)du$$ $$1+\tan^2(u)=\sec^2(u)$$
\begin{align} \int\frac{dx}{\sqrt{a^2-x^2}} & = \int\frac{1}{\sqrt{a^2+a^2\tan^2(u)}}\cdot ai\sec^2(u)du \\ & = ai\int \frac{\sec^2(u)}{\sqrt{a^2(1+\tan^2(u))}}du \\ & = \frac{ai}{a} \int \frac {\sec^2(u)}{\sec(u)}du \\ & = i \int \sec(u)du \\ & =i \ln \lvert \sec(u)+\tan(u)\rvert + C \\ & = i \ln \bigl| \frac {\sqrt{x^2-a^2}}{ai} + \frac {x}{ai} \bigr| + C \\ & = i \ln \bigl| \frac {-i\sqrt{x^2-a^2}}{a} - \frac {ix}{a} \bigr| + C \\ & = i \ln \bigl| -i\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C\\ & = i \Big[ \ln \lvert -i \rvert + \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| \Big] + C\\ & = i \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C \end{align}
