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 :)
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.