# [SOLVED]Same integration but e = 1

#### dwsmith

##### Well-known member
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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 ...