# Thread: Integration Into Arctan

1. ## Integration Into Arctan

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....

Please assist. Thank you

2. Originally Posted by kinkojun

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....

Please assist. Thank you
$\int \frac{1}{3^2 + (2z)^2} \, dz
$

let $u = 2z$

$du = 2 \, dz$

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

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

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

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

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

let $u = 2z$

$du = 2 \, dz$

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

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

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

$\frac{1}{6}\arctan\left(\frac{2z}{3}\right) + C
$
thank you so much!