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[SOLVED] Same integration but e = 1


Well-known member
Feb 1, 2012
Looking for a more elegant solution:
\int_0^{\nu}\frac{d\nu'}{(1 + \cos\nu')^2}.
We can make the substitution $\tan\frac{\nu'}{2} = u$ but I would like another method if possible.


Well-known member
MHB Math Helper
Jan 17, 2013
I think of two ways but I have not tried them , maybe you can try and tell me the complexity :

1- multiply by \(\displaystyle \frac{(1-\cos v')^2}{(1-\cos v')^2}\)

2- Use integration by parts to solve \(\displaystyle \int _0^v \frac{-\sin v'}{-\sin v' \,(1+\cos v')^2}\, dv'\)

For me the second approach seems better ...