Integral of Ldv/(AgL+v²) in inverse tangent form

[btbam91]

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 whenarctan = integral of

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

Help is appreciated.

 

[tiny-tim]

hi btbam91!

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

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

(or go straight to v = √(AgL)tanθ)

 

[HallsofIvy]

Equivalently, divide both numerator and denominator of
$$\frac{ L dv}{AgL+ v^2}$$
by AgL:
$$\frac{\frac{1}{Ag}dv}{1+ \frac{v^2}{AgL}}$$
and then make the substitution
$$u= \frac{v}{\sqrt{AgL}}$$
Because $du= dv/\sqrt{AgL}$ that changes the integral to
$$\sqrt{\frac{L}{Ag}}\int \frac{du}{1+ u^2}$$

 

[btbam91]

Thanks guys! I really appreciate it!