Integral $\int \frac {\sin x} {\sqrt{1-\cos^2 x}} dx$

Question

$\int \frac {\sin x} {\sqrt{1-\cos^2 x}} dx$

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?

Answer 1

$$|\sin^2 x|=\sin^2x, $$ since its always positive. Also notice that you have something like $$\int \frac{-f'(x) }{\sqrt{1-f(x) ^2}}dx, $$ that's arcosine..

Answer 2

Your integrand is just $$\frac{\sin x}{|\sin x|}=\begin{cases}1 & (2k\pi <x < (2k+1) \pi)\\ -1 & ((2k+1)\pi <x< (2k+2) \pi) \end{cases}, \ \ \ k\in \mathbb Z.$$ (See figure below, blue plot.) Therefore the primitive functions are all the functions of the form $$F(x) = \begin{cases}x + C_{a_k} & (2k\pi <x < (2k+1) \pi)\\ -x + C_{b_k} & ((2k+1)\pi <x< (2k+2) \pi) \end{cases},\ \ \ k \in \mathbb Z, $$ (with $C_{a_k}$ and $C_{b_k}$ arbitrary constants) obviously defined for $x \neq k\pi$.

One of the primitive is, e.g. $$\arccos(\cos x), \ \ \ x\neq k\pi$$ (see graph below, red plot) or $$x \cdot \mbox{sign} (\sin x) + \frac{\pi}2,\ \ \ k \neq k\pi$$ (yellow plot).

Answer 3

$$\because \sqrt{1-\cos ^2 x}=\sqrt{\sin ^2 x}=|\sin x|$$ $$ \begin{aligned} \therefore \int \frac{\sin x}{\sqrt{1-\cos ^2 x}} d x & =\int \frac{\sin x}{|\sin x|} d x \\ & =x \operatorname{sgn}(\sin x)+C \end{aligned} $$

Answer 4

Let $u=\cos x$, then $d u=-\sin x d x$ and $$ \begin{aligned} \int \frac{\sin x}{\sqrt{1-\cos ^2 x}} d x & =-\int \frac{d u}{\sqrt{1-u^2}} \\ & =\arccos u+C \\ & =\arccos (\cos x)+C \end{aligned} $$