1. ## Improper Intergrals...

evaluate:

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

after a trig sub I end up here:

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

I'm not sure where to go from here?

This is what I was thinking:

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

but not sure. Any help in finishing this is greatly appreciated!

2. Hello, Got5onIt!

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

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

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

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

3. if you don't want to use the standard form for the answer, you could have used the trig substitution $\displaystyle \frac t2 = \sec \theta$ (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 $\displaystyle \theta$, the limits would not be 2 to 4 anymore)

4. Thanks for the help!

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

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

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

Looks like I was all wrong or going the "unnecessary" route.

5. Originally Posted by Got5onIt
$\displaystyle \int_2^{4} \frac{dt}{t\sqrt{t^2-4}}$
When the discontinuity is visible (of course, if it's in the upper or lower integration limit, not on the interval $\displaystyle [a,b])$, you can avoid the limit procedure, then evaluate. The rest is routine.

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

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