Question

need help with finding the integral

Answer 1

\[\int\limits_{?}^{?} \sqrt(x^2-4)\2x\]

Answer 2

uhh...
Do you mean
\[\Large \int \sqrt{x^2 - 4} \ \text{dx}\]
?

Answer 3

that over 2x

Answer 4

and you got to use trigonometric substitution

Answer 5

Oh... like this
\[\Large \int \frac{\sqrt{x^2-4}}{2x}dx\]?

Answer 6

yes

Answer 7

Looks like you already know how to get started :D
How do you substitute?

Answer 8

so far i have 1/2\[\int\limits_{?}^{?} 2\tan \theta * \tan \theta d\]

Answer 9

back up, lol.
What did you substitute for x?

Answer 10

x= 2sec(theta) dx=2sec(theta)tan(theta) d(theta) and the sqrt(x^2-4)=2tan(theta)

Answer 11

Okay, let's just put that in..
\[\large x = 2\sec(\theta) \\\large dx = 2\sec(\theta)\tan(\theta) d\theta \]
\[\large \sqrt{x^2 -4}= 2\tan(\theta)\]
Making the integral
\[\Large \int \frac{2\tan(\theta)}{4\sec(\theta)}\cdot 2\tan(\theta)\sec(\theta)d\theta\]
So far so good haha

Answer 12

yeah

Answer 13

And it does simplify into you integral.
\[\Large \int 2\tan^2(\theta)d\theta\]
and I assume this is where you got stuck?

Answer 14

yep

Answer 15

i dont know what to do after that

Answer 16

Just hang your head in despair? HAHA
Kidding. Well... I'm assuming you have no difficult integrating \(\sec^2(\theta)\) ?

Answer 17

how do you do that?

Answer 18

How to do what? Integrate \(\sec^2(\theta)\)?
LOL
What's the derivative of \(\tan(\theta)\)? haha

Answer 19

sec^2x

Answer 20

you mean \(\theta\).
And yeah, since the derivative of \(\tan(\theta)\) is \(\sec^2(\theta)\), then it only makes sense that the integral of \(\sec^2(\theta)\) is \(\tan(\theta) + C\)

Understood?

Answer 21

but we dont get a sec^2 theta we get a sec theta

Answer 22

No... we get a \(\tan^2 (\theta)\) haha

Answer 23

This, remember?
\[\Large \int 2\tan^2(\theta)d\theta\]

Or maybe it should be like this instead, to avoid confusion:
\[\Large \color{red}2\int \tan^2(\theta)d\theta\]

Answer 24

Okay, let's go back to trigonometric identities.... specifically, the pythagorean identities...

You CAN express \(\tan^2(\theta)\) in terms of \(\sec(\theta)\). Do you know how?

Answer 25

\[2 \int\limits_{?}^{?} \frac{ \tan^2 \theta }{ 2\sec \theta }\]

Answer 26

so thats the equation we have so far right?

Answer 27

No... let's go back a bit.
\[\Large \int \frac{2\tan(\theta)}{4\sec(\theta)}\cdot 2\tan(\theta)\sec(\theta)d\theta\]

\[\Large \int \frac{\color{blue}{\cancel2}\tan(\theta)}{\color{blue}{\cancel4}\sec(\theta)}\cdot \color{blue}{\cancel2}\tan(\theta)\sec(\theta)d\theta\]
these cancel out.

Answer 28

\[\Large \int \frac{\tan(\theta)}{\cancel{\sec(\theta)}}\cdot \tan(\theta)\cancel{\sec(\theta)}d\theta\]
And these cancel out.

Leaving...
\[\Large \int \tan^2(\theta)\]
Yeah, sorry about that.

Answer 29

oops, I meant \[\Large \int \tan^2(\theta) d\theta\]

Answer 30

okay

Answer 31

Now, as I said, you need to express \(\tan^2(\theta)\) in terms of \(\sec(\theta)\)
Can you do that?

Answer 32

not so sure

Answer 33

Fine. Promise me you'll review your trigonometric identities, and I'll show you how it's done.
Promise? ^^

Answer 34

okay i promise

Answer 35

Okay, Pythagorean identities. One of them is
\[\Large \tan^2(\theta) + 1 = \sec^2(\theta)\]
Which means that
\[\Large \tan^2(\theta) = \sec^2(\theta) - 1\]

Now..., we substitute this into our integral:
\[\Large \int \tan^2(\theta)d\theta = \int \left[\sec^2(\theta) - 1\right]d\theta\]
Can you integrate this now?

Answer 36

yeah i think i can do it

Answer 37

Okay. Then my work here is done? :D

Answer 38

can i let you know when i get the final answer so you can tell me if im right

Answer 39

Sure. I'll just take a shower haha

Answer 40

alright cool

Answer 41

Math usually makes me feel unclean too :p

Answer 42

lol

Answer 43

I can let you know if you answer is good or not if kurt wants to enjoy his shower

Answer 44

LOL Thanks for covering for me then XD

Answer 45

alright

Answer 46

okay i got tan(theta) - theta + C

Answer 47

Ok you just have to put that back in terms of x

Answer 48

okay how?

Answer 49

remember your sub was x/2=sec(theta)