$\int\frac{\sin(x)}{\sqrt{1-\cos^2(x)}}$ Where did I mess up the domain?

Question

So we have $$\int\frac{\sin(x)}{\sqrt{1-\cos^2(x)}}dx$$ My book uses a $u$-substitution with $u=\cos(x)$, $du=-\sin(x)$,and they get $$\int\frac{-du}{\sqrt{1-u^2}}$$ which gives them $\arccos(u)+C=\arccos(\cos(x))+C$

I understand that the domain of $\arccos(x)$ means that $\arccos(\cos(x))$ is only equal to $x$ on $[0,\pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution. $$\int\frac{\sin(x)}{\sqrt{1-\cos^2(x)}}dx$$ $1-\cos^2(x)=\sin^2(x)$ so $$\int\frac{\sin(x)}{\sqrt{1-\cos^2(x)}}dx=\int\frac{\sin(x)}{\sqrt{\sin^2(x)}}=\int\frac{\sin(x)}{\sin(x)}dx=\int{1\cdot dx}$$ So I get $x + C$ as my answer. Where did I mess up?

Answer 1

You messed up because $\sqrt{\sin^2(x)} =\sin x$ is not true for all values of $x$. In fact, it is only true if $\sin(x)\geq0$. If $\sin(x)<0$, then $\sqrt{\sin^2(x)}=-\sin(x)$. In general, $\sqrt{\sin^2(x)} = |\sin x|$.

Answer 2

We have $\frac{sin(x)}{\sqrt{sin^2(x)}}=\frac{sin(x)}{|\sin(x)|}=1 \iff \sin(x) >0$

and $\frac{sin(x)}{\sqrt{sin^2(x)}}=\frac{sin(x)}{|\sin(x)|}=-1 \iff \sin(x) <0$.

Can you take it from here ?

Answer 3

The integrand is a square wave. Its antiderivative is a triangle wave.