1. Integration by Trig Substition

Lower boundary = 2sqrt2

upper boundary = 4

(sqrt(x^2 - 4)) / x

So far i've set x = 2 sectheta

dx = 2 (sec theta) (tan theta)

From here i'm having problems what I should do next

I know I have to place new boundaries.

The new lower boundary is gonna pi / 4

the upper boundary i'm having trouble figuring out. what is the sec theta when it equals 2?

2. Originally Posted by JonathanEyoon
Lower boundary = 2sqrt2

upper boundary = 4

(sqrt(x^2 - 4)) / x

So far i've set x = 2 sectheta

dx = 2 (sec theta) (tan theta)

From here i'm having problems what I should do next

I know I have to place new boundaries.

The new lower boundary is gonna pi / 4

the upper boundary i'm having trouble figuring out. what is the sec theta when it equals 2?
$\displaystyle 4=2\sec(\theta) \iff 4=\frac{2}{\cos(\theta)} \iff \cos(\theta)=\frac{1}{2} \iff \theta =\cos^{-1}\left( \frac{1}{2}\right)=\frac{\pi}{3}$

3. You have the correct substitution.

Subbing in you get:

$\displaystyle \int\frac{\sqrt{(2sec(t))^{2}-4}}{2sec(t)}2sec(t)tan(t)dt$

This all reduces down to:

$\displaystyle 2\int{tan^{2}(t)}dt$

Now, the new limits are $\displaystyle 2\sqrt{2}=2sec(t)\rightarrow{t=\frac{\pi}{4}}$

$\displaystyle 4=2sec(t)\rightarrow{t=\frac{\pi}{3}}$

So, we have:

$\displaystyle 2\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}tan^{2}(t)dt$

Can you finish now?.

4. I appreciate it fellas. I'll take it from here and let you know if I have any other troubles. Thanks

5. Actually I am gonna need some further help. How would I integrate tan^2(x)?

I'll go ahead and take a guess and say. Break up the tan^2(X)

sin^2(x) * (1 / cos^2(x))

U - substitution?

6. Originally Posted by JonathanEyoon
Actually I am gonna need some further help. How would I integrate tan^2(x)?

I'll go ahead and take a guess and say. Break up the tan^2(X)

sin^2(x) * (1 / cos^2(x))

Integration by parts?
Try this trig identity

$\displaystyle \tan^2(x)+1=\sec^2(x) \iff \tan^{2}(x)=\sec^2(x)-1$

remember that $\displaystyle \frac{d}{dx}\tan(x)=\sec^2(x)$

7. I tried to use that identity but unfortunately i'm just getting more confused. Can you give me a hint on what the next step would be? Thanks

Never mind. I figured it out.!! *sigh* I hate it when I make careless mistakes!

8. Originally Posted by JonathanEyoon
I tried to use that identity but unfortunately i'm just getting more confused. Can you give me a hint on what the next step would be? Thanks

$\displaystyle \int \tan^2(x)dx=\int(\sec^2(x)-1)dx$

from my hint before that $\displaystyle \frac{d}{dx}\tan(x)=\sec^2(x)$ this implies that $\displaystyle \int \sec^2(x)dx=\tan(x)$ so we get...

$\displaystyle \int \tan^2(x)dx=\int(\sec^2(x)-1)dx=\tan(x)-x$

9. Hello,

Without using sec..

$\displaystyle \tan^2x=\underbrace{1+\tan^2x}_{\text{derivative of tan x}}-1$

--> $\displaystyle \int \tan^2x \ dx=\int 1+\tan^2x \ dx-\int dx$

Just another way to do it