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integrals with roots
- Thread starter CalcGuy
- Start date
tutor_joel
Junior Member
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- Feb 6, 2010
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- 157
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?
-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?
tutor_joel
Junior Member
- Joined
- Feb 6, 2010
- Messages
- 157
Thanks for the feedback, glad I could help
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
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- 1,577
\(\displaystyle \int_{-2}^{-9}\frac{\sqrt{x^{2}-1}}{x^{3}}dx \ = \ .367424, \ (trusty \ TI-89)\)
\(\displaystyle Let \ x \ = \ sec(t), \ \implies \ dx \ = \ sec(t)tan(t)dt\)
\(\displaystyle Hence, \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{sec^{2}(t)-1}]sec(t)tan(t)}{sec^{3}(t)}dt\)
\(\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{tan^{2}(t)]}sec(t)tan(t)}{sec^{3}(t)}dt\)
\(\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{|tan(t)|tan(t)}{sec^{2}(t)}dt, \ |tan(t)| \ = \ -tan(t), \ 2nd \ quadrant\)
\(\displaystyle = \ -\int_{2\pi/3}^{arcsin(1/9)+\pi/2}sin^{2}(t)dt \ = \ -1/2\int_{2\pi/3}^{arcsin(1/9)+\pi/2}(1-cos(2t))dt\)
\(\displaystyle = \ -1/2\bigg[t-\frac{sin(2t)}{2}\bigg]_{2\pi/3}^{arcsin(1/9)+\pi/2} \ = \ -1/2[t-sin(t)cos(t)]_{2\pi/3}^{arcsin(1/9)+\pi/2}\)
\(\displaystyle = \ -1/2[arcsin(1/9)+\pi/2-(4\sqrt5/9)(-1/9)-(2\pi/3-(\sqrt3/2)(-1/2))] \ = \ .367424 \ QED\)
\(\displaystyle Let \ x \ = \ sec(t), \ \implies \ dx \ = \ sec(t)tan(t)dt\)
\(\displaystyle Hence, \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{sec^{2}(t)-1}]sec(t)tan(t)}{sec^{3}(t)}dt\)
\(\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{[\sqrt{tan^{2}(t)]}sec(t)tan(t)}{sec^{3}(t)}dt\)
\(\displaystyle = \ \int_{2\pi/3}^{arcsin(1/9)+\pi/2}\frac{|tan(t)|tan(t)}{sec^{2}(t)}dt, \ |tan(t)| \ = \ -tan(t), \ 2nd \ quadrant\)
\(\displaystyle = \ -\int_{2\pi/3}^{arcsin(1/9)+\pi/2}sin^{2}(t)dt \ = \ -1/2\int_{2\pi/3}^{arcsin(1/9)+\pi/2}(1-cos(2t))dt\)
\(\displaystyle = \ -1/2\bigg[t-\frac{sin(2t)}{2}\bigg]_{2\pi/3}^{arcsin(1/9)+\pi/2} \ = \ -1/2[t-sin(t)cos(t)]_{2\pi/3}^{arcsin(1/9)+\pi/2}\)
\(\displaystyle = \ -1/2[arcsin(1/9)+\pi/2-(4\sqrt5/9)(-1/9)-(2\pi/3-(\sqrt3/2)(-1/2))] \ = \ .367424 \ QED\)