Solving Trigonometric Integral Using Trig Substitution and Limits of Integration

[3.141592654]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Using trig. sub, I have sec \theta=t

dt=sec \theta tan \theta d \theta

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

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

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

=\int\frac{tan \theta d \theta}{(sec \theta)^2\sqrt{(tan \theta)^2}}

=\int\frac{tan \theta d \theta}{(sec \theta)^2tan \theta}

=\int\frac{d \theta}{(sec \theta)^2}

=\int\cos \theta^2 d\theta

=\int\frac{1+cos \theta}{2} d\theta

=\int\frac{1}{2}+\frac{cos \theta}{2} d\theta

=\frac{1}{2}\int d \theta +\frac{1}{2}\int cos \theta d \theta

=\frac{1}{2} \theta | +\frac{1}{2}sin \theta |

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

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

I need to get to the answer: \frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}. 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?

 

[LCKurtz]

\cos^2\theta = \frac{1 + \cos{(2 \theta})}{2}

 

[3.141592654]

That was a silly mistake.

Alright, so:

\int (cos \theta)^2 d\theta

=\int \frac{1+cos(2\theta)}{2} d\theta

=\frac{1}{2}\int d\theta+\frac{1}{2}\int cos(2\theta)d\theta

=\frac{1}{2}\theta| +\frac{1}{2}\frac{sin(2\theta)}{2}|

=\frac{1}{2}\theta| +\frac{1}{4}sin(2\theta)|

=\frac{1}{2}\theta| +\frac{1}{4}(2sin\theta cos\theta)|

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t^2})|

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(1-\frac{1}{t^2})|

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

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

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

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

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

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
<br /> \frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}<br />?

Thanks.

 

[ideasrule]

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t^2})|

=\frac{1}{2}arcsin (t)| +\frac{1}{2}(1-\frac{1}{t^2})|

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

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

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

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

=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{3}{8}-\frac{1}{4}
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.