$$f: \mathbb R \to \mathbb R, \: f(x) = \frac{\sin ^{2n}\left(x\right)}{\cos ^{2n}\left(x\right)+\sin ^{2n}\left(x\right)},\:n\:\in \mathbb{N}^* fixed$$
1)Find the principal period of the function.
In order to solve this, I have applied the formula $\sin^2{x} = \frac{1-\cos{2x}}{2}$ and I have stated that the period $T = \pi$. Does this make sense ?
2) The function $f + c$ where $c \in \mathbb R$ has a periodic primitive if and only if the value of $c = $ ?
I don't really know how to approach this problem, all I know is that the period of $f$ is $k\pi$...
The answer for 2) is $\frac{-1}{2}$
Surely $\pi$ is a period for $f$, because $\sin(x+\pi)=-\sin x$ and $\cos(x+\pi)=-\cos x$.
What you should prove is that this is the minimal period. If we consider \begin{align} f'(x)&= \frac{ 2n\sin^{2n-1}x\cos x(\sin^{2n}x+\cos^{2n}x)- \sin^{2n}x(2n\sin^{2n-1}x\cos x-2n\cos^{2n-1}x\sin x)} {(\sin^{2n}x+\cos^{2n}x)^2} \\ &=\frac{2n\sin^{2n-1}x\cos^{2n-1}x}{(\sin^{2n}x+\cos^{2n}x)^2} \end{align} we see that $f$ has its first point of maximum at $\pi/2$.
In order that a periodic function has periodic derivative, you need that the integral over a period is $0$. You can note that the maximum of $f$ is $1$ (at $\pi/2$) and the minimum is at $0$. Also $f(\pi-x)=f(x)$, for $x\in[0,\pi]$. Then…