# Trigonometric integral

## Homework Statement

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

## 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 $$[\sqrt{2}, 2]$$

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

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?

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member
$$\cos^2\theta = \frac{1 + \cos{(2 \theta})}{2}$$

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
$$\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}$$?

Thanks.

ideasrule
Homework Helper
$$=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|$$

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

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

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.