Why don't we draw modulus bars when we open 'under root' in indefinite integration? Is there a way we can ignore absolute value bars in indefinite integration. Below is the solution of a problem that confused me and not just this problem, I recently noticed that all the problem and questions that I've been doing , I was omitting modulus when opening roots.
$$\int{\sqrt{1+\sin x}\,\,dx}$$
Multiplying and dividing by $\sqrt{1-\sin x}\,\,$ , we get,
$$\int{\frac{\sqrt{(\cos)^2 x}}{\sqrt{1-\sin x}}\,\,dx}$$
Now in the solution, they opened the root of Cos square x and wrote just Cos x and not |cos x|.
Please explain me why is it so that in every indefinite problem I am finding that modulus bars are omitted.
And if it is not so, then correct me where am I wrong , also in just another case of my confusion.
If we let, $$\sin x = t$$ then what would be  $ \cos x$
$$\cos x = \pm \sqrt{1-t^2}$$ But they usually omit -ve sign and take only $ \sqrt{1-t^2}$
These two are the most confusing parts to me.  I  really need these doubts get cleared
Please help me as I am new to calculus and facing problems :)
 A: It is not correct to not take care of the modulus.
A correct procedure, I think, could be the following:
\begin{align}
\int\sqrt{1+\sin x}\,dx 
    &= \int\frac{\sqrt{1-\sin^2 x}}{\sqrt{1-\sin x}}\,dx = \\
    &= \int\frac{\sqrt{\cos^2x}}{\sqrt{1-\sin x}}\,dx = \\
    &= \int\frac{|\cos x|}{\sqrt{1-\sin x}}\,dx =(*).
\end{align}
Observe that, after the first step, the new integrand function is not defined in the points $\pi/2+2k\pi,$ but can be extended there by continuity, so this is not a problem for the integrability.
To simplify further elaboration, let's write
$$
|\cos x| = \varepsilon\cos x,
$$
where
$$
\varepsilon=\mathrm{sign}(\cos x)=
\begin{cases}
+1 & \text{if}\ \exists k\in\mathbb{Z}:\ -\pi/2+2k\pi<x<+\pi/2+2k\pi \\
-1 & \text{if}\ \exists k\in\mathbb{Z}:\ +\pi/2+2k\pi<x<+3\pi/2+2k\pi
\end{cases}
$$
The integral becomes
\begin{align}
(*) 
    &= \varepsilon\int\frac{\cos x}{\sqrt{1-\sin x}}\,dx = \\
    &= \varepsilon\int\frac{1}{\sqrt{1-t}}\,dt = \\
    &= -2\varepsilon\sqrt{1-t} + c = \\
    &= -2\varepsilon\sqrt{1-\sin x} + c = \\
    &= -2\varepsilon\frac{\sqrt{1-\sin^2x}}{\sqrt{1+\sin x}} + c = \\
    &= -2\varepsilon\frac{\sqrt{\cos^2x}}{\sqrt{1+\sin x}} + c = \\
    &= -2\varepsilon\frac{|\cos x|}{\sqrt{1+\sin x}} + c = \\
    &= -2\frac{\cos x}{\sqrt{1+\sin x}} + c.
\end{align}
where the last four steps are devoted to make the cosine function to appear again, so that we can substitute back $\varepsilon|\cos x|$ with $\cos x.$
The result so obtained is valid for each values of $x$, excluded the values that make the denominator equal to $0$, where there are jump discontinuities:

To overcome these discontinuities, we can add a different integration constant on each interval in which the function is continuous, and define the primitive as
$$
-2\frac{\cos x}{\sqrt{1+\sin x}} + 4\sqrt{2}\lfloor\frac{x+\pi/2}{2\pi}\rfloor + c.
$$
the graphic representation been

Finally, an animation to show the change from the first to the second graphics:


Edit
The relation
$$
\cos x = \sqrt{1-\sin^2x}
$$
is not correct written this way. Commonly it is written as
$$
\cos x = \pm\sqrt{1-\sin^2x},
$$
where the sign is defined the same way as above, i.e it can be written as
$$
\cos x = \varepsilon\sqrt{1-\sin^2x},
$$
with $\varepsilon$ defined previously. The advantage of this writing is that it ease calculations, while the $\pm$ is more difficult to take car of. Alternatively, it could be written
$$
|\cos x| = \sqrt{1-\sin^2x}.
$$
The analogous relation
$$
\sin x = \pm\sqrt{1-\cos^2x}
$$
can also be written as
$$
\sin x = \varepsilon'\sqrt{1-\cos^2x}
$$
or
$$
|\sin x| = \sqrt{1-\cos^2x}
$$
but here $\varepsilon'$ has a different definition, being related to the sign of $\sin x$, and not $\cos x$. We have
$$
\varepsilon'=\mathrm{sign}(\sin x)=
\begin{cases}
+1 & \text{if}\ \exists k\in\mathbb{Z}:\ 2k\pi<x<\pi+2k\pi, \\
-1 & \text{if}\ \exists k\in\mathbb{Z}:\ \pi+2k\pi<x<2\pi+2k\pi.
\end{cases}
$$
A last note, the choice of symbols $\varepsilon,\varepsilon'$ is my personal choice, and in no way standard notation. Also, don't know if there can be other ways more efficient to address the problem of the sign.
