$\sqrt{x^2}=|x|$ or $\sqrt{x^2}=x$ in an indefinite integral

Question

Question:

Find the following integral: $\int{\sqrt{1+\cos(x)}}dx$

My attempt:

$$\int{\sqrt{1+\cos(x)}}dx$$

$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$

$$=\int{\sqrt{2}\ \left|\cos\frac{x}{2}\right|}dx$$

$$=\sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx\tag4$$

$$=\sqrt{2}\int{\pm\cos\frac{x}{2}}dx$$

$$=\pm\sqrt{2}\int{\cos\frac{x}{2}}dx$$

$$=\pm2\sqrt{2}\sin\frac{x}{2}+C$$

My book's attempt:

$$\int{\sqrt{1+\cos(x)}}dx$$

$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$

$$=\int{\sqrt{2}\cos\frac{x}{2}}dx$$

$$=\sqrt{2}\int{\cos\frac{x}{2}}dx$$

$$=2\sqrt{2}\sin\frac{x}{2}+C$$

Basically, my book didn't put $\pm$ sign, while I did. My book did this essentially: $\sqrt{x^2}=x$, while I did this: $\sqrt{x^2}=|x|$. Is my process more correct?

Answer 1

1. $(y=|x|)$ is universally equivalent not to $(y=\pm x),$ but to $(y=\pm x \;\;\text{and}\;\; y\ge0),$ so $|x|$ and $\pm x$ are not mutually substitutable.
2. $$\int_0^{2\pi}{\left|\cos\frac{x}{2}\right|}\,\mathrm dx = 4,$$ but $$\pm\int_0^{2\pi}{\cos\frac{x}{2}}\,\mathrm dx=0= \int_0^{2\pi}{\cos\frac{x}{2}}\,\mathrm dx.$$

The correct way to continue from line $(4)$ is to split up the integral such that each of the integrand's $x$-intercepts is a limit of integration.