$\int \frac {\sin x} {\sqrt{1-\cos^2 x}} dx$
My attempt: Obviously, we can call $u=\cos x \implies du=-\sin x dx$ and rest can be done.
However, I thought at the beginning the following: I can rewrite integrant as $$\int \frac {\sin x dx} {\sqrt{1-\cos^2 x}}=\int \frac {\sin x dx}{|\sin^2 x|}$$
But I couldn't continue because I don't know how to seperate absolute value because of indefinite integral. So, is there any mistake about my idea?
Best Answer
$$|\sin^2 x|=\sin^2x, $$ since its always positive. Also notice that you have something like $$\int \frac{-f'(x) }{\sqrt{1-f(x) ^2}}dx, $$ that's arcosine..