# 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 ?