# Integration Into Arctan

• Mar 6th 2010, 10:33 AM
kinkojun
Integration Into Arctan
http://i277.photobucket.com/albums/k...8/DSC00031.jpg

I couldn't understand this as i have tried using the formula... it leads me to (1/3)*arctan(2z/3) instead of 1/6 in front....

• Mar 6th 2010, 10:49 AM
skeeter
Quote:

Originally Posted by kinkojun
http://i277.photobucket.com/albums/k...8/DSC00031.jpg

I couldn't understand this as i have tried using the formula... it leads me to (1/3)*arctan(2z/3) instead of 1/6 in front....

$\displaystyle \int \frac{1}{3^2 + (2z)^2} \, dz$

let $\displaystyle u = 2z$

$\displaystyle du = 2 \, dz$

$\displaystyle \frac{1}{2} \int \frac{2}{3^2 + (2z)^2} \, dz$

$\displaystyle \frac{1}{2} \int \frac{1}{3^2 + u^2} \, du$

$\displaystyle \frac{1}{2} \cdot \frac{1}{3}\arctan\left(\frac{u}{3}\right) + C$

$\displaystyle \frac{1}{6}\arctan\left(\frac{2z}{3}\right) + C$
• Mar 6th 2010, 11:37 AM
kinkojun
Quote:

Originally Posted by skeeter
$\displaystyle \int \frac{1}{3^2 + (2z)^2} \, dz$

let $\displaystyle u = 2z$

$\displaystyle du = 2 \, dz$

$\displaystyle \frac{1}{2} \int \frac{2}{3^2 + (2z)^2} \, dz$

$\displaystyle \frac{1}{2} \int \frac{1}{3^2 + u^2} \, du$

$\displaystyle \frac{1}{2} \cdot \frac{1}{3}\arctan\left(\frac{u}{3}\right) + C$

$\displaystyle \frac{1}{6}\arctan\left(\frac{2z}{3}\right) + C$

thank you so much!