Will the graphs of equivalent indefinite integrals always look identical? I came across the following indefinite integral
$$\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x \space  (0 \leq x \leq \pi )$$
I attempted to solve it as follows:
$$ u = 1 - \sin\left(2x\right) \implies \sin\left(2x\right) = 1- u$$ 
$$\implies x = \dfrac{\arcsin\left ( 1 - u\right)}{2} \implies \mathrm{d}x = \dfrac{-\mathrm{d}u }{2 \sqrt{1- (1-u)^2} }$$
$$\therefore \textrm{ we have }\space \dfrac{-1}{2} \int \dfrac{ \sqrt{u} } {\sqrt { 1 - (1 - u)^2}} \space \mathrm{d}u = \dfrac{-1}{2} \int \sqrt {\dfrac{ u } { 1 - (1 -2u + u^2)}} \space \mathrm{d}u $$
$$  =  \dfrac{-1}{2} \int \sqrt {\dfrac{u} {u(2-u)}} \space \mathrm{d}u = \dfrac{-1}{2} \int \sqrt {\dfrac{ 1 } { 2-u}} \space \mathrm{d}u $$
Let $z = 2 - u$. Then $\mathrm{d}z = -\mathrm{d}u$
$$\implies \dfrac{1}{2} \int \sqrt {\dfrac{ 1 } {z}} \space \mathrm{d}z = \dfrac{1}{2} \int z^{-1/2}= \sqrt{z} + C = \sqrt{2-u} + C = \sqrt{1 + \sin\left(2x\right)} + C $$
I ran the problem through the site Integral Calculator and got
$$\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x = -\sin{x} - \cos{x} + C$$ 
which only coincides with my solution if I take its absolute value. Refer to the image below: 
$\sqrt{1 + \sin\left(2x\right)} + C $ is in orange, whereas $-\sin{x} - \cos{x} + C$ is in purple. In the diagram, $C = 0$. The shaded region is the given range of $x$ values : $0 \leq x \leq \pi$.

Now, from what I know, the graphs of equivalent indefinite integrals should look similar and should differ only by a constant, which means they will coincide after a translation transformation. However, it seems not to be the case here. Hence my question; will the graphs of equivalent indefinite integrals always look similar, or perhaps there is a mistake in my calculations?
In addition, the answer that is given in the problem book is 
$$2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \mathrm{sgn} \space t \cdot \left\{ \cos{ \dfrac{t}{\pi} } - \cos{t}\right\} \\ \mathrm{ where } \space t = x - \dfrac{\pi}{4} \space \mathrm{ and } \left[ \cdot \right] \textrm{ is the integer part of the expression inside } $$
This further confused me because now I don't know the right answer and where I went wrong. Any help is appreciated.
 A: 
Note that the integration is over a periodic function with periodic non-differentiable points as shown in the plot. Follow the steps below to perform such indefinite integration.
$$I=\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x 
= \int {\sqrt{(\sin{x} - \cos{x})^2 }}\space \mathrm{d}x $$
$$=  \int |\sin{x} - \cos{x}|  \mathrm{d}x =\sqrt 2 \int |\sin(x-\frac{\pi}{4})| \mathrm{d}x = \sqrt 2\int |\sin t|dt$$
where the variable change $t=x-\frac \pi4$ is made in the last step. Note that the last expression is a positive periodic function with periodity $\pi$ and its repeated integral value in each periodic interval is given by
$$A=\int_0^\pi |\sin t| dx = 2$$
Then, for $t\ge 0$, reeexpress the integral as,
$$I =\sqrt 2 \int_0^{[\frac t\pi]} |\sin s| ds+\sqrt 2 \int_{[\frac t\pi]}^t \sin s ds + C $$ 
$$=\sqrt 2 A\left[\frac t\pi\right]  - \sqrt 2 \cos s|_{[\frac t\pi]}^t + C$$
$$=2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \left( \cos{ \dfrac{t}{\pi} } - \cos{t}\right) +C $$
The result for $t<0$ can be derived similarly. The combined integral result then reads,
$$I = 2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \mathrm{sgn} \space t \cdot \left(\cos{ \dfrac{t}{\pi} } - \cos{t}\right) +C $$
