Integral $\int\sqrt{1-\cos2x}~dx=$? So here is the problem I'm working with
$$\int\sqrt{1-\cos2x}~dx$$
I'm assuming that I'll need to use the trig identity
$2\sin^2x=1-\cos2x$ . But where do I go from there?
$$\int\sqrt{2\sin^2x}~dx$$
Do I split the $\sqrt{2}$ and the $\sqrt{\sin^2x}$ so that I have $$\int\sqrt{2}\sin x~dx$$ and do integration by parts?
 A: Yes, you can simplify as you did at the end, but no need for integration by parts! Recall, $\sqrt 2$ is merely a constant!
$$\int \sqrt 2 \sin x \,dx = \sqrt 2 \int \sin x \,dx = -\sqrt 2 \cos x + C$$
You did the "hardest part" by recognizing the trigonometric identity here. The rest is simply knowing that $\int \sin x = -\cos x + C$
A: $\newcommand{\+}{^{\dagger}}%
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Integrate $\verts{\sin\pars{x}}$, for example, as follows.
With $x > 0$:
\begin{align}
&\int_{0}^{x}\verts{\sin\pars{t}}\,\dd t =
x\verts{\sin\pars{x}} - \int_{0}^{x}t\ \sgn\pars{\sin\pars{t}}\cos\pars{t}\,\dd t
\\[3mm]&=
x\verts{\sin\pars{x}} - \int_{0}^{x}\sgn\pars{\sin\pars{t}}\phi'\pars{t}\,\dd t
\quad\mbox{where}\quad\phi\pars{x} \equiv \int_{0}^{x}t\cos\pars{t}\,\dd t
\\[3mm]
&\int_{0}^{x}\verts{\sin\pars{t}}\,\dd t = x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\phi\pars{x} + \int_{0}^{x}\phi\pars{t}\bracks{2\delta\pars{\sin\pars{t}}\cos\pars{t}}\,\dd t
\\[3mm]&=
x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\phi\pars{x} + 2\int_{0}^{x}\phi\pars{t}\sum_{n = 0}^{n\pi \leq x}\delta\pars{t - n\pi}\cos\pars{t}\,\dd t
\\[3mm]&=
x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\phi\pars{x} +
2\sum_{n = 0}^{n\pi \leq x}\pars{-1}^{n}\phi\pars{n\pi}
\end{align}
Also
\begin{align}
\phi\pars{x}&=x\sin\pars{x} - \int_{0}^{x}\sin\pars{t}\,\dd t
=
x\sin\pars{x} + \cos\pars{x} - 1
\end{align}
\begin{align}
\int_{0}^{x}\verts{\sin\pars{t}}\,\dd t
=
\sgn\pars{\sin\pars{x}}\bracks{1 - \cos\pars{x}} + 2\sum_{n = 0}^{n\pi \leq x}\pars{-1}^{n}\bracks{\cos\pars{n\pi} - 1}
\end{align}
$$\color{#0000ff}{\large%
\int_{0}^{x}\verts{\sin\pars{t}}\,\dd t
=
\sgn\pars{\sin\pars{x}}\bracks{1 - \cos\pars{x}} + 4\sum_{n = 0}^{\pars{2n + 1}\pi \leq x}1}
$$
