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Trigonometric integral

  • #1

Homework Statement



[tex]\int\frac{1}{t^3\sqrt{t^2-1}}dt[/tex] with limits of integration [tex][\sqrt{2}, 2][/tex]

Homework Equations





The Attempt at a Solution



Using trig. sub, I have [tex]sec \theta=t[/tex]

[tex]dt=sec \theta tan \theta d \theta[/tex]

[tex]\int\frac{1}{t^3\sqrt{t^2-1}}dt[/tex] with limits of integration [tex][\sqrt{2}, 2][/tex]

[tex]=\int\frac{sec \theta tan\theta d \theta}{(sec \theta)^3\sqrt{(sec \theta)^2-1}}[/tex]

[tex]=\int\frac{tan \theta d \theta}{(sec \theta)^2\sqrt{(sec \theta)^2-1}}[/tex]

[tex]=\int\frac{tan \theta d \theta}{(sec \theta)^2\sqrt{(tan \theta)^2}}[/tex]

[tex]=\int\frac{tan \theta d \theta}{(sec \theta)^2tan \theta}[/tex]

[tex]=\int\frac{d \theta}{(sec \theta)^2}[/tex]

[tex]=\int\cos \theta^2 d\theta[/tex]

[tex]=\int\frac{1+cos \theta}{2} d\theta[/tex]

[tex]=\int\frac{1}{2}+\frac{cos \theta}{2} d\theta[/tex]

[tex]=\frac{1}{2}\int d \theta +\frac{1}{2}\int cos \theta d \theta[/tex]

[tex]=\frac{1}{2} \theta | +\frac{1}{2}sin \theta |[/tex]

[tex]=\frac{1}{2}arcsec t +\frac{1}{2}t[/tex] with limits [tex][\sqrt{2},
2][/tex]

[tex]=\frac{1}{2}[arcsec (2)-arcsec (\sqrt{2})]+\frac{1}{2}[2-\sqrt{2}][/tex]

I need to get to the answer: [tex]\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}[/tex]. I don't see anything wrong up to this point so I guess my question is more of an algebra question, but how could I arrive at the stated answer?
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
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[tex]\cos^2\theta = \frac{1 + \cos{(2 \theta})}{2}[/tex]
 
  • #3
That was a silly mistake.

Alright, so:

[tex]\int (cos \theta)^2 d\theta[/tex]

[tex]=\int \frac{1+cos(2\theta)}{2} d\theta[/tex]

[tex]=\frac{1}{2}\int d\theta+\frac{1}{2}\int cos(2\theta)d\theta[/tex]

[tex]=\frac{1}{2}\theta| +\frac{1}{2}\frac{sin(2\theta)}{2}|[/tex]

[tex]=\frac{1}{2}\theta| +\frac{1}{4}sin(2\theta)|[/tex]

[tex]=\frac{1}{2}\theta| +\frac{1}{4}(2sin\theta cos\theta)|[/tex]

[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|[/tex]

[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t^2})|[/tex]

[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(1-\frac{1}{t^2})|[/tex]

with limits of integration [tex][\sqrt{2}, 2][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{2^2})-(1-\frac{1}{\sqrt{2}^2})[/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2})][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2}})][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{3}{8}-\frac{1}{4}[/tex]

I'm not sure if this is correct, but how do I deal with the arcsin values since they aren't in the domain of the standard arcsin function? Lastly, how would I algebraically convert this to
[tex]
\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}
[/tex]?

Thanks.
 
  • #4
ideasrule
Homework Helper
2,266
0
[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|[/tex]

[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t^2})|[/tex]

[tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(1-\frac{1}{t^2})|[/tex]

with limits of integration [tex][\sqrt{2}, 2][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{2^2})-(1-\frac{1}{\sqrt{2}^2})[/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2})][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2}})][/tex]

[tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{3}{8}-\frac{1}{4}[/tex]
Thanks.
You don't have to go through all this drama. You know that t=sec(theta), so cos(theta)=1/t. Just use this to convert the limits of integration to theta.
 

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