- #1
geft
- 148
- 0
Given
[tex]R = \sqrt{d^2 + z^2}[/tex]
and
[tex]\cos \alpha = \frac{d}{R}[/tex]
find
[tex]\int \frac{1}{d^2 + z^2}dz[/tex]
The answer given is
[tex]\frac{1}{d}\tan^{-1}(\frac{z}{d})[/tex]
I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.
[tex]R = \sqrt{d^2 + z^2}[/tex]
and
[tex]\cos \alpha = \frac{d}{R}[/tex]
find
[tex]\int \frac{1}{d^2 + z^2}dz[/tex]
The answer given is
[tex]\frac{1}{d}\tan^{-1}(\frac{z}{d})[/tex]
I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.