How can i evaluate the integral : Pi/2 $(1-3) Sqrt (t^2-1)dt

[Riazy]

Homework Statement

The question scanned

Homework Equations

The Attempt at a Solution

I don't really know what to do

 

[craig16]

Probably try trig sub, with t equal to sec(theta)
don't forget to tack on a sec(theta)tan(theta) when you change dt to dtheta

Edit: I meant sec(theta)tan(theta) not tan^2(theta)

so it becomes

Integral of sec(theta)tan$$^{2}$$(theta) dtheta

 

[HallsofIvy]

$sin^2(t)+ cos^2(t)=1$ so that, dividing by $cos^2(t)$, $tan^2(t)+ 1= sec^2(t)$. That is,
$$\sqrt{tan^2(t)- 1}= sec(t)$$

And, of course, if x= tan(t), $dx= sec^2(t)dt$.

 

[pergradus]

$sin^2(t)+ cos^2(t)=1$ so that, dividing by $cos^2(t)$, $tan^2(t)+ 1= sec^2(t)$. That is,
$$\sqrt{tan^2(t)- 1}= sec(t)$$

And, of course, if x= tan(t), $dx= sec^2(t)dt$.

This is incorrect. The correct substitution is

$t = sec(x)$ and then the radical becomes $\sqrt{sec^2(x)- 1}= tan(x)$

and $dt/dx = sec(x)tan(x)$. The integral then becomes:

$$\int{tan(x)sec(x)tan(x)dx} = \int{sec(x)tan^2(x)dx} = \int{sec(x)-sec^3(x)dx}$$

edit - I don't know why my latex is all screwed up (I ****ING HATE LATEX) but hopefully he can understand what i wrote

 

[dextercioby]

So

$$I = \frac{\pi}{2}\int\limits_{1}^{3} \sqrt{x^2 - 1} \, dx$$.

Well, let $x=\cosh t$ first, compute the new limits and the new integral and then at the end use the double angle formula for $\sinh t$

$$\sinh^2 t = \frac{1}{2}\left(\cosh 2t - 1\right)$$