# Which is correct ?

• Nov 6th 2005, 09:08 AM
kippy
Which is correct ?
Hi.

(Let, Int=integral and Sqrt=square root and In=natural log)

In a math book I found that

Int( Sqrt(Z^2-A^2) )
= (A^2)/2 * ( (Z*Sqrt(Z^2-A^2))/(A^2) - In((Z+sqrt(z^2-A^2))/A) )
= (Z*Sqrt(Z^2-A^2))/2 - ( (A^2)/2 * In((Z+sqrt(z^2-A^2))/A) )

but with my TI-89 titanium calculator the same integral gave

Int( Sqrt(Z^2-A^2) )
= (Z*Sqrt(Z^2-A^2))/2 - ( (A^2)/2 * In(Z+sqrt(z^2-A^2)) )

After some working out on paper, I found my calculator gave the wrong result. I not too sure why this is. Can anyone tell me which is correct ?

Thanks
• Nov 6th 2005, 11:17 AM
Jameson
Which variable is this respect to?

$\displaystyle \int \sqrt{z^2-a^2}da$ is much different from $\displaystyle \int \sqrt{z^2-a^2}dz$

Jameson
• Nov 6th 2005, 01:21 PM
rgep
You can check an alleged indefinite integral (anti-derivative) by differentiating ...
• Nov 6th 2005, 03:13 PM
kippy
ah... sorry, it's with repect to z.

thanks
• Nov 6th 2005, 03:22 PM
Jameson
$\displaystyle \int \sqrt{x^2-a^2}dx=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2\ln(x+\sqrt{x^2-a^2})}{2}$