Can You Simplify This Trigonometric Integral?

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SUMMARY

The discussion focuses on simplifying the trigonometric integral $$\int_0^{\nu}\frac{d\nu'}{(1 + \cos\nu')^2}$$. Two alternative methods are proposed: first, multiplying by $$\frac{(1-\cos v')^2}{(1-\cos v')^2}$$, and second, using integration by parts on the integral $$\int _0^v \frac{-\sin v'}{-\sin v' \,(1+\cos v')^2}\, dv'$$. The second method is favored for its potential elegance and effectiveness in solving the integral.

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Dustinsfl
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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.
 
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I think of two ways but I have not tried them , maybe you can try and tell me the complexity :

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

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

For me the second approach seems better ...
 

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