# Improper Intergrals...

• November 6th 2007, 08:12 PM
Got5onIt
Improper Intergrals...
evaluate:

$\int_2^{4} \frac{dt}{t\sqrt{t^2-4}}$

after a trig sub I end up here:

$\int_2^{4} \frac{d\theta}{tan\theta}$

I'm not sure where to go from here?

This is what I was thinking:

$\lim_{b\rightarrow 2} \left[ \int \frac{d\theta}{tan\theta} \right]_{b}^{4}$

but not sure. Any help in finishing this is greatly appreciated!
• November 6th 2007, 08:38 PM
Soroban
Hello, Got5onIt!

Quote:

Evaluate: . $\int_2^{4} \frac{dt}{t\sqrt{t^2-4}}$

After a trig sub I end up here: . $\int_2^{4} \frac{d\theta}{\tan\theta}$ . . . . no

This is a standard form: . $\int\frac{dx}{x\sqrt{x^2-a^2}} \;=\;\frac{1}{a}\text{ arcsec}\,\frac{x}{a} + C$

So we have: . $\frac{1}{2}\text{ arcsec}\,\frac{t}{2}\,\bigg]^4_2 \;=\;\frac{1}{2}\bigg[\text{arcsec}\,2 - \text{arcsec}\,1\bigg] \;=\;\frac{1}{2}\bigg[\frac{\pi}{3} - 0\bigg] \;=\;\frac{\pi}{6}$

• November 6th 2007, 08:41 PM
Jhevon
if you don't want to use the standard form for the answer, you could have used the trig substitution $\frac t2 = \sec \theta$ (after some algebraic manipulation), and you would arrive at the answer

(by the way, when you change the variable, the limits usually changes with it, in this case, when you went from t to $\theta$, the limits would not be 2 to 4 anymore)
• November 6th 2007, 09:07 PM
Got5onIt
Thanks for the help!

I used the trig sub $t= 2sec\theta$

and $dt= 2\sec\theta tan\theta d\theta$

and $\sqrt{t^2-4} = \sqrt{4(sec^2\theta-1} = 2tan\theta$

Looks like I was all wrong or going the "unnecessary" route.:D
• November 7th 2007, 04:08 AM
Krizalid
Quote:

Originally Posted by Got5onIt
$\int_2^{4} \frac{dt}{t\sqrt{t^2-4}}$

When the discontinuity is visible (of course, if it's in the upper or lower integration limit, not on the interval $[a,b])$, you can avoid the limit procedure, then evaluate. The rest is routine.

--

Why do we have to apply trig. sub.? Before that, I suggest to set $u=\sqrt{t^2-4}\implies du=\frac t{\sqrt{t^2-4}}\,dt.$

Since $u^2+4=t^2,$ you'll have an easy integral to evaluate, and the conclusion follows.