<s>[INST] Absolute values in $\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$ in my math class we were given a list of indefinite integrals, and one of them was:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$
My working:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int \frac{dx}{(x+2)\sqrt{(x+2)^2-1}}$$
Then I used the substitution $x+2=\sec t$ to get:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int \frac{\tan t}{|\tan t|}dt= t\,\text{sgn}\, (\tan t)+C...$$
Then I checked the answer sheet, and this is what they did:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int dt=t+C=\text{arcsec}(x+2)+C$$
What I don't understand is, why are they allowed to say $\sqrt{\sec^2 t-1}=\tan t?$ I tried to put some values in and I have found that:
$$\int_{\sec \left(\frac{8}{5}\right)-2}^{\sec \left(\frac{9}{5}\right)-2} \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}<0$$ but according to the answer sheet I would get $\dfrac{1}{5}$
My answer looks wrong, I would be happy if someone could explain what the problem is, and also why we are allowed to simplify like they did.
 [/INST] The answer of the book looks wrong. Since the principal range of $\sec x$ is:


*

*$\left[ 0; \frac{\pi}{2} \right)$ if $x \ge 1$, and on this domain, $\tan t$ is positive.

*$\left( \frac{\pi}{2} ; \pi \right]$ if $x \le -1$, whereas on this domain, $\tan t$ is negative.


So, in fact, the solution to that integral should be split in to 2 different parts:
$$\int\limits \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}} = \left\{ \begin{array}{ll} -\mbox{arcsec}(x + 2) + C_1 &, \mbox{for }x < -3 \\
\mbox{arcsec}(x + 2) + C_2 &, \mbox{for }x > -1 \end{array} \right.$$
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