integrals with roots

[CalcGuy]

ok ive been stuck on this for a few hours... and im sure its simple im just not seeing it...

see attachment...

i dont understand how they got the new limits?

 

[tutor_joel]

It's pretty straight forward. You can see where the inverse sec (-9) comes from right? since it's simply a straight substitution. the 2pi/3 however comes from the unit circle. So, if

-2 = sec(t), then cos(t) = -1/2 right? Which is in the second quad of the unit circle, and what is the value of the angle? 120 degrees or 2pi/3

Make sense?

 

[CalcGuy]

thank you... i tried that but drew my triangle wrong on the unit circle... i knew better and told my self to pay attention and make sure i didnt do it but i did it anyway...

 

[tutor_joel]

Thanks for the feedback, glad I could help

 

[BigGlenntheHeavy]

$\displaystyle \int_{-2}^{-9}\frac{\sqrt{x^{2}-1}}{x^{3}}dx \ = \ .367424, \ (trusty \ TI-89)$

$\displaystyle Let \ x \ = \ sec(t), \ \implies \ dx \ = \ sec(t)tan(t)dt$

$\displaystyle Hence, \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{sec^{2}(t)-1}]sec(t)tan(t)}{sec^{3}(t)}dt$

$\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{tan^{2}(t)]}sec(t)tan(t)}{sec^{3}(t)}dt$

$\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{|tan(t)|tan(t)}{sec^{2}(t)}dt, \ |tan(t)| \ = \ -tan(t), \ 2nd \ quadrant$

$\displaystyle = \ -\int_{2\pi/3}^{arcsin(1/9)+\pi/2}sin^{2}(t)dt \ = \ -1/2\int_{2\pi/3}^{arcsin(1/9)+\pi/2}(1-cos(2t))dt$

$\displaystyle = \ -1/2\bigg[t-\frac{sin(2t)}{2}\bigg]_{2\pi/3}^{arcsin(1/9)+\pi/2} \ = \ -1/2[t-sin(t)cos(t)]_{2\pi/3}^{arcsin(1/9)+\pi/2}$

$\displaystyle = \ -1/2[arcsin(1/9)+\pi/2-(4\sqrt5/9)(-1/9)-(2\pi/3-(\sqrt3/2)(-1/2))] \ = \ .367424 \ QED$