- #1
yungman
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I don't get this answer, this is my work:
Find [tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} [/tex]
Let [tex] tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s [/tex]
[tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = \int^a_0 sec \;\theta \;d\theta = ln| sec \;\theta + tan \;\theta| = ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right |^a_0 [/tex]
But from the book:
[tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |^a_0 [/tex]
Can anyone help?
Thanks
Find [tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} [/tex]
Let [tex] tan \;\theta = \frac z s \;\Rightarrow dz = s \;sec^2\;\theta \;d \theta, \; sec \;\theta = \frac {\sqrt{s^2 + z^2}} s [/tex]
[tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = \int^a_0 sec \;\theta \;d\theta = ln| sec \;\theta + tan \;\theta| = ln \left | \frac {\sqrt{s^2+z^2} + z}{s} \right |^a_0 [/tex]
But from the book:
[tex]\int^a_0 \frac {dz}{\sqrt{ s^2+z^2 }} = ln | \sqrt{s^2+z^2} + z |^a_0 [/tex]
Can anyone help?
Thanks
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