$\sqrt{x^2}=|x|$ 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.

Answer 2

Your solution goes further than the book's solution in terms of correctness. You have two issues. The first is that changing $\left|\cos\frac{x}{2}\right|$ to $\pm\cos\frac{x}{2}$ only creates a loss of information. Moreover, the $\pm$ sign itself depends on that value of $\cos\frac{x}{2}$, so it makes no sense to pull out the $\pm$ sign here, even if you did have it.

Integrating over absolute value signs is tricky. I'll use the definite integral to give an answer to the indefinite integral.

$$ \sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx = \sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C $$

In the case of a definite integral, you have to split your domain into intervals on a case by case basis.

Case 1: $x>\pi$. In this case, we have \begin{align} \sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C &= \sqrt{2}\int_{0}^{\pi} \left|\cos\frac{u}{2}\right| du \\ & \quad+ \left( \sqrt{2}\int_{\pi}^{\pi+2\pi} \left|\cos\frac{u}{2}\right| du + \cdots + \sqrt{2}\int_{\pi+(n-1)2\pi}^{\pi+n2\pi} \left|\cos\frac{u}{2}\right| du \right) \\ & \quad+ \sqrt{2}\int_{\pi+n2\pi}^{x} \left|\cos\frac{u}{2}\right| du + C \qquad\qquad \end{align} where $n = \left\lfloor \dfrac{x-\pi}{2\pi}\right\rfloor$.

We look at the various definite integrals separately. First, note that $\cos\frac{u}{2}$ is nonnegative for all $u\in (0, \pi)$. This means we have $\left|\cos\frac{u}{2}\right| = \cos\frac{u}{2}$ on that interval. Hence $$ \sqrt{2}\int_{0}^{\pi} \left|\cos\frac{u}{2}\right| du = \sqrt{2}\int_{0}^{\pi} \cos\frac{u}{2} du = 2\sqrt{2}. $$

Next, using periodicity of cosine, we can deduce that $$ \sqrt{2}\int_{\pi+(k-1)2\pi}^{\pi+k2\pi} \left|\cos\frac{u}{2}\right| du = \sqrt{2}\int_{-\pi}^{\pi} \cos\frac{u}{2} du = 4\sqrt{2} $$ for each integer $k$.

Lastly, we need to do the last integral, which is going to give us a function of $x$. Again, using properties of cosine, we can deduce that \begin{align} \sqrt{2}\int_{\pi+n2\pi}^{x} \left|\cos\frac{u}{2}\right| du &= \sqrt{2}\int_{-\pi}^{x - 2(n+1)\pi} \cos\frac{u}{2} du = \sqrt{2}\cdot 2\sin\left(\frac{u}{2}\right) \Big|_{-\pi}^{x-(n+1)2\pi} \\ &= 2\sqrt{2} \left[ \sin\left(\frac{x-(n+1)2\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right] \\ &= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}-(n+1)\pi\right) - (-1) \right] \\ &= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}\right)\cos(-(n+1)\pi) + \sin(-(n+1)\pi)\cos\left(\frac{x}{2}\right) + 1 \right] \\ &= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}\right)(-1)^{n+1} + 0 + 1 \right] \\ &= 2\sqrt{2} \left[ 1 + (-1)^{n+1}\sin\left(\frac{x}{2}\right) \right] \end{align}

Going all the way back to the beginning of Case 1, and putting this all together, we have \begin{align} \sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C &= 2\sqrt{2} + \underbrace{(4\sqrt{2} + \cdots + 4\sqrt{2})}_{n\text{ terms}} + 2\sqrt{2}[1 + (-1)^{n+1}\sin\left(\frac{x}{2}\right)] + C \\[1.2ex] &= 2\sqrt{2} + n(4\sqrt{2}) + 2\sqrt{2} + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C \\[1.6ex] &= (n+1)\cdot 4\sqrt{2} + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C \end{align} Thus, for $x > \pi$ we have $$ \sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx = 4\sqrt{2}(n+1) + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C $$ where $n = \left\lfloor \dfrac{x-\pi}{2\pi}\right\rfloor$.

The solution might look weird, but when I graph it in desmos for $C=0$ I get a continuous function.

Cases for $0\le x\le \pi$, $-\pi\le x<0$, and $x<-\pi$ are similar. I'll leave these other cases to you :)

The reason why we have all these cases and the reason why we need to split our definite integral over various intervals is because the absolute value function is actually a piecewise function that is itself handled by cases.

Answer 3

Before a radical sign imho...

the non inclusion of $\pm$ in the last step is a convention. It is assumed that it is understood by implication , so not so necessary even to mention.

This practice is annoying when integral evaluations are done..

Omission of sign in differential equation is a misleading step that needs to be altogether avoided.

Your procedure is more correct.If you graph them they are seen as reflections on either side of x-axis, each in its own right, so to say.

Answer 4

Note that $\sqrt{t^2}=\text{sgn}(t)\cdot t$. Then, piecewise integrate

\begin{align} \int\sqrt{1+\cos x}\>dx= &\>\sqrt2\int\sqrt{\cos^2\frac x2}\>dx =\>\sqrt2 \>\text{sgn}(\cos \frac x2)\int\cos \frac x2 \>dx\\ = &\>2\frac{ \sqrt{2\cos^2\frac x2}}{\cos \frac x2}\sin \frac x2 =\>2\sqrt{1+\cos x}\>\tan \frac x2 \end{align} A global ante-derivative can be concatenated as $$ \int\sqrt{1+\cos x}\>dx= 2\sqrt{1+\cos x}\>\tan \frac x2 +4\sqrt2\lfloor \frac {x+\pi}{2\pi}\rfloor $$