$
\int \frac x {\sqrt {1-x^2}}dx
$
I was attempting to solve this integral, and it would appear the solution to it is $-\sqrt{1 -x^2}+C$. When I attempted to solve it, however, I attempted to let $x = \sin\theta$, making $dx=\cos{\theta}d{\theta}$
$
\int \frac {\sin\theta} {\cos^2\theta}\cos\theta{d\theta} = \int \tan \theta d\theta = \ln|\sec\theta| + C = \ln{\frac 1 {\sqrt {1 – x^2}}}+C = -{\frac 1 2}\ln|1-x^2|+C
$
I don't quite understand why this is incorrect. Now, I do understand that what I had to do to get the correct solution is to let $u=\sqrt {1-x^2}$, and everything else works out. But can someone explain where I went wrong with my attempt?
Best Answer
After the substitution you should get $\int \frac{\sin\theta}{\cos\theta}\cos\theta\;d\theta$. You probably forgot the square root.