Question

I am trying to solve this problem: http://www.wolframalpha.com/input/?i=int%3A+sqrt%28x^2-4%29%2F2x

Answer 1

Because of Right Triangles rules (see: http://imgur.com/PgoZi) I have the following identities:
2tanx=sqrt((x^2)
2secx=x
^(derivative): 2tanxsecx=dx

I now have all the parts of the initial equation and can substitute accordingly:
1/2int:(2tanx/2secx)(2tanxsecx) = (1/2)int:sinx2tanxsecx = int:sin^2x/cos^2x
= int:tan^2x

Next, I attempt to plug in sqrt(x^2-4)/2 for tan in the integral right above:
int:(sqrt(x^2-4)/2)^2 = (1/4)int:x^2-4 = (1/4)(((x^3)/3)-4x) + c

Answer 2

I am unsure whether I have made an error as my answer does not match the one presented on Wolfram Alpha at all.. could anyone PLEASE help me?

Answer 3

*2tan=sqrt((x^2)-4)

Answer 4

Some of your text is difficult to work with, would you mind using the equation editor in the future? Because you didn't notate them, did you make certain that you kept careful track of your bounds of integration?

Answer 5

I am sorry, There wouldn't be a problem using the equation editor if we could use in when typing the inital question or if we had the ability to edit! >.<

Answer 6

There are no bounds of integration

Answer 7

Time for bed, Ill be back in the morning!

Answer 8

\[\int\limits_{}^{}\frac{\sqrt{x^2-4}}{2x}dx\]
\[x=2\sec(\theta) \]
\[dx=2\sec(\theta)\tan(\theta)d\theta\]
\[\int\limits{}{}\frac{4\tan^2(\theta)\sec(\theta)}{4\sec(\theta)}d\theta\]
\[\int\limits{}{}\tan^2(\theta)d\theta\]
\[\tan(\theta)-\theta\]
-----
\[\sqrt{(\frac{x}{2})^2-1}=\tan(\theta)=\frac{\sqrt{x-4}}{2}\]
\[\theta=\sec^{-1}(\frac{x}{2})\]

^ See step 2
\[\frac{\sqrt{x-4}}{2}-\sec^{-1}(\frac{x}{2})+C\]
Is what I got, and it agrees with Wolfram's result (it gives it as an alternate form).

Answer 9

Thanks so much!