Prove that $f(x)$ is periodic with period $2\pi$ if $n$ is odd. Prove that for even $n$ the function is a sum of linear function and periodic function: $$ f(x) = \int_0^x\sin^n(t)dt $$
I started with considering the case when $n$ is odd. I tried to use a proof by induction. Let $n = 1$ be a base case: $$ f_1(x) = \left.\int_0^x\sin(t)dt = -\cos(t)\right|_0^x = 1-\cos x $$ Which is indeed periodic with the period $2\pi$. Suppose this holds for $n = 2k+1$: $$ f_{2k+1}(x) = f_{2k+1}(x+2\pi) = \int_0^x\sin^{2k+1}(t)dt $$
Consider the case for $n = 2k+3$: $$ \begin{align} f_{2k+3}(x) &= \int_0^x\sin^{2k+3}(t)dt \\ &= \int_0^x\sin^{2k+1}(t)\sin^2(t)dt \\ &= \int_0^x\sin^{2k+1}(t)(1-\cos^2t)dt \\ &= \int_0^x\sin^{2k+1}(t)dt - \int_0^x\cos^2tdt \\ &= \int_0^x\sin^{2k+1}(t)dt - \left.\left({t\over 2} + {\sin(2t)\over 4}\right)\right|_0^x \\ &= \int_0^x\sin^{2k+1}(t)dt - \left({x\over 2} + {\sin(2x)\over 4}\right) \end{align} $$ Which is obviously not periodic. I believe the problem here is that induction requires that we make a single step for $n$ while I have added $2$.
So I tried another approach. Let $n$ be odd, then: $$ \begin{align} f(x) &= \int_0^x\sin^{2k+1}(t)dt \\ &= \int_0^x\sin^{2k}(t)\sin(t)dt \\ &= -\int_0^x(1-\cos^2)^kd(\cos(t)) \end{align} $$ The problem is I don't see how to proceed from here.
I would appreciate it if someone can show a way to prove the statement above for odd and even $n$. Thanks!
$$I(x+2\pi)=\int_0^{2\pi+x}\sin^n(t)\,dt=\int_0^{2\pi}\sin^n(t)\,dt+\int_0^{x}\sin^n(t)\,dt=J+I(x).$$
When $n$ is odd, $J=0$ and the function $I$ is periodic.
When $n$ is even,
$$\hat I(x):=\int_0^x\left(f(x)-\dfrac J{2\pi}\right)\,dx=I(x)-\frac J{2\pi}x$$ is such that
$$\hat I(x+2\pi)=\hat I(x).$$