Find all values of $p$ for which the function $f(x) = \frac{1}{(1-\sin x)^{1/6}}$ belongs to $L^p[0,2\pi]$ I'd be thankful if somebody can help me with this.
I want to find all the values of $p$ for which the function
${\rm f}\left(x\right) = {1 \over \left[1-\sin\left(x\right)\right]^{1/6}}$ belongs to $L^{p}\left[0,2\pi\right]$.
My thoughts so far:
By showing that the Riemann-integral diverges I can not conclude anything about Lebesgue-integrability and If the Riemann integral meant something, the problem is that the function is not bounded in $\left[0,2\pi\right]$ so it is not Riemann-integrable. What method can I use to prove that
$\int^{2\pi}_{0}\left\vert{\rm f}\left(x\right)\right\vert^{\,p}\,{\rm d}x$ converges or diverges? I tried to use Holder's inequality, but that does not seem to be of any help.

Essentially, the region that causes trouble is around $\pi/2$. Hence, $$I = \int_{\pi/2-\delta_1}^{\pi/2+\delta_2} \dfrac{dx}{(1-\sin(x))^{p/6}} = \int_{-\delta_1}^{\delta_2} \dfrac{dx}{(1-\cos(x))^{p/6}} = \dfrac1{2^{p/6}}\int_{-\delta_1}^{\delta_2} \dfrac{dx}{\vert \sin(x/2) \vert^{p/3}}$$
Hence,
$$I \sim {2^{p/6}}\int_{-\delta_1}^{\delta_2} \dfrac{dx}{\vert x\vert^{p/3}}$$
which is integrable if $p/3 < 1 \implies p <3$.
