Evaluating a common integral for $\arcsin(x)$ using a different method

Question

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}

Answer 1

First, we have for complex $u$

$$\int \sec(u)\,du=\log(\sec(u)+\tan(u))+C$$

Next, we note that for $x=ia\tan(u)$ and $a\ge x$

$$\begin{align} i\log(\sec(u)+\tan(u))+C&=i\log\left(\frac{\sqrt{x^2-a^2}}{ia}+\frac{x}{ia}\right)+C\\\\ &=i\log(x+i\sqrt{a^2-x^2})+C'\\\\ &=-\arctan\left(\frac{\sqrt{a^2-x^2}}{x}\right)+C''\\\\ &=\arctan\left(\frac{x}{\sqrt{a^2-x^2}}\right)+C'''\\\\ &=\arcsin(x/a)+C'''' \end{align}$$

And we are done.

Answer 2

Hint:

Note that $$y = \sin(x) = \frac{e^{ix}-e^{-ix}}{2} = \frac{e^{i2x}-1}{2e^{ix}}.$$ Consequently, $$(e^{ix})^2+2y\cdot e^{ix}-1=0 \implies e^{ix} = -y\pm\sqrt{y^2+1} \implies x = ?$$