# integrate using substitution x = 2sec@

• Apr 2nd 2010, 03:47 AM
differentiate
integrate using substitution x = 2sec@
Integrate
sqrt (x^2 -4) using the substitution x = 2sec@

x = 2sec@
dx = 2sec@tan@

sqrt (4sec^2 @ -4) x 2sec@tan@ d@
= 4tan^2 @ sec@

and now I'm stuck.

Thanks
• Apr 2nd 2010, 03:51 AM
Sudharaka
Quote:

Originally Posted by differentiate
Integrate
sqrt (x^2 -4) using the substitution x = 2sec@

x = 2sec@
dx = 2sec@tan@

sqrt (4sec^2 @ -4) x 2sec@tan@ d@
= 4tan^2 @ sec@

and now I'm stuck.

Thanks

Ted had written answer before I did. So I deleted mine. Anyway this integration could be easily done if you use x=2sin@.

Hope this helps you.
• Apr 2nd 2010, 03:51 AM
Ted
Quote:

Originally Posted by differentiate
Integrate
sqrt (x^2 -4) using the substitution x = 2sec@

x = 2sec@
dx = 2sec@tan@

sqrt (4sec^2 @ -4) x 2sec@tan@ d@
= 4tan^2 @ sec@

and now I'm stuck.

Thanks

$\displaystyle \int tan^2(\theta) sec(\theta) \, d\theta=\int (sec^2(\theta)-1)sec(\theta) \, d\theta=\int sec^3(\theta) \, d\theta - \int sec(\theta) \, d\theta$

the first one is a well-known integral and you should know how to integrate the second one