How to compute
$$ \int_{0}^{2\pi}\dfrac{1-\cos(t)}{\biggl(\dfrac{5}{4}-\cos(t)\biggr)^{\dfrac{3}{2}}} dt $$
I'm interested in more ways of computing this integral.
My thoughts:
I'm tired to use Bioche rules :
Let $x=\operatorname{tg}(\frac{t}{2})$ then $$\cos(t)=\dfrac{1-x^{2}}{1+x^{2}},\qquad dt=\dfrac{2}{1+x^{2}}dx$$
\begin{align} \int_{0}^{2\pi}\dfrac{1-\cos(t)}{\biggl(\dfrac{5}{4}-\cos(t)\biggr)^{\dfrac{3}{2}}} dt&=\int_{0}^{2\pi}\dfrac{1-\dfrac{1-x^{2}}{1+x^{2}}}{\biggl(\dfrac{5}{4}-\dfrac{1-x^{2}}{1+x^{2}}\biggr)^{\dfrac{3}{2}}} \dfrac{2}{1+x^{2}}dx \\ &=\int_{0}^{2\pi}\dfrac{\frac{2x^2}{x^2+1}}{\left(\frac{9x^2+1}{4\left(x^2+1\right)}\right)^{\frac{3}{2}}}\dfrac{2}{1+x^{2}}dx\\ &=\int_{0}^{2\pi}\frac{4x^2}{\left(\frac{9x^2+1}{4\left(x^2+1\right)}\right)^{\frac{3}{2}}\left(x^2+1\right)^2}dx \end{align}
First we can see that the integral is twice the integral over $[0,\pi]$ and further using $\cos t=1-2\sin^2(t/2)$ and substitution $t=2x$ we get $$I=64\int_{0}^{\pi/2}\frac{\sin^2x}{(1+8\sin^2x)^{3/2}}\,dx=64\int_{0}^{\pi/2}\frac{\sin^2x}{(9-8\cos^2x)^{3/2}}\,dx$$ and this further leads to $$I=\frac{64}{27}\int_{0}^{\pi/2}\frac{1-\sin^2x}{(1-k^2\sin^2x)^{3/2}}\,dx$$ where $k^2=8/9$. It should now be anyone's guess that we are going to have to deal with elliptic integrals. The following uses the notation $$K(k) =\int_{0}^{\pi /2}\frac{dx}{\sqrt{1-k^2\sin^2x}},\,E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx$$ We can now write $$1-\sin^2x=\frac{k^2-1+1-k^2\sin^2x}{k^2}$$ and hence $$I=\frac{64(k^2-1)}{27k^2}\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx+\frac{64}{27k^2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}=\frac{8}{3}K(k)-\frac{8}{27}\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx$$ For the integral on right hand side note that $$\frac{d} {dx} \frac{\sin x\cos x} {\sqrt{1-k^2\sin^2x}}=\frac{\sqrt{1-k^2\sin^2x}} {k^2}-\frac{1-k^2}{k^2}\cdot(1-k^2\sin^2x)^{-3/2}$$ and thus integrating the above we get $$\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx=9E(k)$$ and therefore we have the desired integral $$I=\frac{8}{3}(K(k)-E(k))$$