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$$f: \mathbb R \to \mathbb R, \: f(x) = \frac{\sin ^{2n}\left(x\right)}{\cos ^{2n}\left(x\right)+\sin ^{2n}\left(x\right)},\:n\:\in \mathbb{N}^* fixed$$

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 $\frac{-1}{2}$

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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…

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Hint: if $F$ is a periodic primitive of $f(x)+c$ with period $T$, then $$F(T+s)-F(s) = \int_{s}^{T+s} (f(x)+c)dx = 0.$$ Now you need to find $T$ and then find $c$.

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    Have you not simply restated the question? – 2017-02-26
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    Why is that true? I know that $\int _s^{s+T}f\left(x\right)dx=\int _0^T\:f\left(x\right)dx$ not $0$ – 2017-02-26
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    @Liviu that is why you need that constant $c$. – 2017-02-26
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    @Dr.MV I wouldn't say so. After all, I hinted the approach with integrals. – 2017-02-26
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    @TZakrevskiy So, you've given away the fact that a primitive is an integral? – 2017-02-26