integration – Trigonometric Substitution Absolute value issue

Evaluate $ , displaystyle int _{0}^{4} frac{1}{(2x+8), sqrt{x(x+8)}}, dx. $

$My work:-$
by completing the square and substitution i.e. $displaystyle left(begin{array}{rl}x+4 & = 4sec (theta )\ dx & = 4sec theta tan theta , dtheta end{array}right) qquad$
$Rightarrow displaystyle int frac{4sec theta tan theta , dtheta }{2(4sec (theta ))( |4tan (theta )|) }$

$Rightarrow displaystyle frac{1}{8} int frac{tan theta , dtheta }{|tan (theta )|}$

now because my limits are positive so $sec theta geq 0 $ and $sec theta $ is positive in $ Ist $ and $ IVth $ Quadrant. At this stage i have 2 options either i consider $ Ist $ Quadrant and take postive $|tan theta| = tan theta $ or i consider $ IVth $ Quadrant where $ |tan theta| = -tan theta$

So when in 1st quadrant i.e. $ |tan theta| = tan theta ,$ $ 0 geq theta geq pi/2 $ i get

$Rightarrow displaystyle frac{1}{8} theta +C$

$Rightarrow displaystyle frac{1}{8} mathrm{arcsec}left(frac{x+4}{4}right)+C$

$Rightarrow displaystyle left. frac{1}{8} mathrm{arcsec}left(frac{x+4}{4}right), right|_{x=0}^{x=4}$

$Rightarrow displaystyle frac{1}{8} left(mathrm{arcsec}(2)-mathrm{arcsec}(1)right)$

Now if i consider 4th quadrant i.e. $ |tan theta| = -tan theta ,$ $ 3 pi /2 geq theta geq 2pi $ i get

$Rightarrow displaystyle -frac{1}{8} theta +C$

$Rightarrow displaystyle -frac{1}{8} mathrm{arcsec}left(frac{x+4}{4}right)+C$

$Rightarrow displaystyle left. -frac{1}{8} mathrm{arcsec}left(frac{x+4}{4}right), right|_{x=0}^{x=4}$

$Rightarrow displaystyle -frac{1}{8} left(mathrm{arcsec}(2)-mathrm{arcsec}(1)right)$

So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you’ll get the same answer. so why i m getting Negative answer ?