I am looking for the antiderivative $$\int \frac{\cos^n(x)dx}{1+\cos(x)}$$
at $ (0,\frac{\pi}{2})$.
i used the substitution $$t=\tan(\frac{x}{2})$$ to get $$\int \Bigl(\frac{1-t^2}{1+t^2}\Bigr)^ndt$$
I can't see how to continue and i think there could be a simpler way to do it. Thanks in advance for time.
Let $ n $ be a positive integer, using geometric series formula, we have for any $ x\in\mathbb{R} $ :\begin{aligned}\sum_{k=0}^{n-1}{\left(-1\right)^{k}\cos^{k}{x}}=\frac{1-\left(-1\right)^{n}\cos^{n}{x}}{1+\cos{x}}\end{aligned} Thus : \begin{aligned}\frac{\cos^{n}{x}}{1+\cos{x}}=\frac{\left(-1\right)^{n}}{1+\cos{x}}-\sum_{k=0}^{n-1}{\left(-1\right)^{n-k}\cos^{k}{x}}\end{aligned}
Hence : \begin{aligned}\int{\frac{\cos^{n}{x}}{1+\cos{x}}\,\mathrm{d}x}=\left(-1\right)^{n}\int{\frac{\mathrm{d}x}{1+\cos{x}}}-\sum_{k=0}^{n-1}{\left(-1\right)^{n-k}\int{\cos^{k}{x}\,\mathrm{d}x}}\end{aligned}
Using the fact that $ \cos^{k}{x}=\frac{1}{2^{k}}\sum\limits_{p=0}^{k}{\binom{k}{p}\cos{\left(\left(k-2p\right)x\right)}} $, and that $ 1+\cos{x}=2\cos^{2}{\left(\frac{x}{2}\right)} $, we get : \begin{aligned}\int{\frac{\cos^{n}{x}}{1+\cos{x}}\,\mathrm{d}x}&=\left(-1\right)^{n}\int{\frac{\mathrm{d}x}{2\cos^{2}{\left(\frac{x}{2}\right)}}}-\sum_{k=0}^{n-1}{\sum_{p=0}^{k}{\left(-1\right)^{n-k}\binom{k}{p}\frac{1}{2^{k}}\int{\cos{\left(\left(k-2p\right)x\right)}\,\mathrm{d}x}}}\\ &=\left(-1\right)^{n}\tan{\left(\frac{x}{2}\right)}-\sum_{k=0}^{n-1}{\sum_{p=0}^{k}{\left(-1\right)^{n-k}\binom{k}{p}\frac{f_{k,p}\left(x\right)}{2^{k}}}}+C\end{aligned}
Where $ f_{k,p}\left(x\right)=\left\lbrace\begin{aligned}\frac{\sin{\left(\left(k-2p\right)x\right)}}{k-2p}\ \ \ \ &\textrm{If }k\neq 2p\\ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\textrm{If }k=2p\end{aligned}\right. \cdot $