Trigonometric Substitution Absolute value issue

Question

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

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

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

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\ 2\pi\ $ 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 ?

Answer 1

If $\theta$ lies in 4th quadrant, you need to consider the values of $\text{arcsec}$ in fourth quadrant(as the inverse trig functions are multivalued).

$\text{arcsec }(2) = \theta_1 \Rightarrow \theta_1 = \dfrac{5\pi}3$

$\text{arcsec} (1) = \theta_2 \Rightarrow \theta_2 = 2\pi$

So, $-\dfrac{1}{8}(\theta_1-\theta_2) = -\dfrac{1}{8}\left(\dfrac{-\pi}{3}\right) = \dfrac{\pi}{24}$

This is same as the result from 1st quadrant. $\dfrac18(\text{arcsec }(2)-\text{arcsec }(1)) = \dfrac18\left(\dfrac\pi3-0\right) = \dfrac{\pi}{24}$

Answer 2

HINT

I propose another way to tackle this integral so that you can compare both methods.

Notice that $x(x+8) = x^{2} + 8x = (x + 4)^{2} - 16$.

Hence, if we make the substitution $4\cosh(z) = x + 4$, we arrive at \begin{align*} \int\frac{\mathrm{d}x}{(2x+8)\sqrt{x(x+8)}} & = \int\frac{4\sinh(z)}{8\cosh(z)\sqrt{16\cosh^{2}(z) - 16}}\mathrm{d}z\\\\ & = \frac{1}{8}\int\frac{\mathrm{d}z}{\cosh(z)}\\\\ & = \frac{1}{8}\int\frac{\cosh(z)}{\cosh^{2}(z)}\mathrm{d}z\\\\ & = \frac{1}{8}\int\frac{\cosh(z)}{\sinh^{2}(z) + 1}\mathrm{d}z \end{align*}

where the absolute value was omitted because the function $\sinh(z)$ is positive whenever $z\geq 0$.

In the last expression, we can make the change of variable $w = \sinh(z)$, whence it results that \begin{align*} \int\frac{\mathrm{d}x}{(2x+8)\sqrt{x(x+8)}} = \frac{\arctan(w)}{8} + c \end{align*}

Now it remains to apply the integration limits.

Can you take it from here?