Can someone explain this inverse tan integral for me please?

  • Thread starter btbam91
  • Start date
  • #1
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integral of
(Ldv)/(AgL+v^2) from [v0,0]

I'm supposed to get (...)arctan(...)

where (...) are 2 different quantities.
I'm confused on how to get arctan out of this integral when


arctan = integral of

(1*dv)/(1+v^2) from [0,x]

Help is appreciated.
 

Answers and Replies

  • #2
tiny-tim
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hi btbam91! :smile:

(try using the X2 icon just above the Reply box :wink:)

scale it down … substitute v = x√(AgL) :wink:

(or go straight to v = √(AgL)tanθ)
 
  • #3
HallsofIvy
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Homework Helper
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Equivalently, divide both numerator and denominator of
[tex]\frac{ L dv}{AgL+ v^2}[/tex]
by AgL:
[tex]\frac{\frac{1}{Ag}dv}{1+ \frac{v^2}{AgL}}[/tex]
and then make the substitution
[tex]u= \frac{v}{\sqrt{AgL}}[/tex]
Because [itex]du= dv/\sqrt{AgL}[/itex] that changes the integral to
[tex]\sqrt{\frac{L}{Ag}}\int \frac{du}{1+ u^2}[/tex]
 
  • #4
91
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Thanks guys! I really appreciate it!
 

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