Etydhv

When solving an integral such as $displaystyleintfracdxsqrtx^2+4$, you eventually end up with

$$ lnlvertsectheta+tanthetarvert+C.$$

The next step is to rewrite this in terms of $x$. My book does the following: $x=2tantheta$, so by drawing a triangle, it can be seen that $x$ is the opposite side, $2$ the adjacent one, and finally, $sqrt2^2+x^2$ the hypotenuse. Therefore, $$sectheta=fracsqrt2^2+x^22.$$ The problem that I see however is that $sqrt2^2+x^2$ is the magnitude of the hypotenuse, so this has lost a possible negative sign since it is not necessarily true that

$$ sectheta = frac2.$$

How can this be omitted?

When you make the substitution $x=2tan theta$, you have to be careful to specify the domain of $theta$: the substitution is only valid if $theta$ has a small enough domain for $tan theta$ to be continuous. The simplest possible choice of domain is probably $-fracpi2 < theta < fracpi2$. Note that the range of $2tan theta$ on this domain is the entire real line, so taking $theta$ in this domain doesn't lose any generality.

But when $-fracpi2 < theta < fracpi2$, we always have $sec theta > 0$. So in fact, if you make this choice of domain, it is always true that $2 sec theta=sqrtx^2+4$, without any sign issues.

(If you chose a different domain for $theta$, you might find that $2sectheta = -sqrtx^2+4$ instead. It's a useful exercise to think about why this will not change the eventual value of the integral. In any case, the sign of $sec theta$ is constant on every domain where $tan theta$ is continuous.)

Let's work through it from the beginning
$$x=2tantheta, dx=2sec^2theta dtheta.\
int fracdxsqrtx^2+4=intfrac2sec^2theta dthetasqrt4tan^2theta+4=intfrac2sec^2theta dtheta=int|sectheta|dtheta=intsectheta dtheta\
=ln(tantheta+sectheta)+C.$$

Where the modulus sign can be removed because we can assume $thetain(-fracpi2,fracpi2)$ to get all the values of $x$.

Again, when using $2sectheta=pmsqrtx^2+4$, we know that $thetain(-fracpi2,fracpi2)$, so the minus sign can be ignored.

If you try to use other ranges of $theta$, the negative sign(if any) in the above steps should cancel out, and you can get the same result.

Is this perhaps a possible solution?

$$intfracdx+2sqrtx^2+2^2=intfracdx+2sqrtsec^2(theta)=intfrac2sec^2(theta)2mid sec(theta)middtheta=intmidsec(theta)mid dtheta >0 $$