Derive a recurrence relation for $(a_n)=\int_0^{{\pi / 3}}(\cos{x})^{-n}dx$ I am trying to derive a recurrence relation for
$$(a_n)=\int_0^{{\pi\over 3}}(\cos{x})^{-n}dx$$
$n=0,1,2,\dots$
It is pretty obvious $a_0={\pi \over 3}$. I get stuck on $a_1=\int_0^{{\pi / 3}}{1\over\cos{x}}dx$. I am able to transform it to $a_1=\int_0^{{\sqrt{3} / 2}}{1\over1-y^2}dy$ but not further. Also, this is still quite far from a recurrence equation. Do you have any suggestions?
 A: Hint: Don't use that substitution, instead use integration by parts with $u=(\cos x)^{-(n-2)}$ to get a recurrence between $a_n$ and $a_{n-2}$.
A: hint
for $a_1$, use the substitution
$$\tan (\frac x2)=t $$
with
$$dx=2\frac {dt}{1+t^2} $$
and
$$\cos (x)=\frac {1-t^2}{1+t^2} $$
it becomes
$$\int_0^{\frac {1}{\sqrt {3}}}\frac {2dt}{1-t^2} $$
now use
$$\frac{2}{1-t^2}=\frac {1}{1-t}+\frac{1}{1+t} $$
A: One can just power through and discover 
\begin{align}
a_{n} &= \int_{0}^{\pi/3} \left(\frac{1}{\cos(x)}\right)^{n} \, dx \\
&= \left[ \sin(x) \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{n+1}{2}; \frac{3}{2}; \sin^{2}(x) \right) \right]_{0}^{\pi/3} \\
&= \frac{\sqrt{3}}{2} \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{n+1}{2}; \frac{3}{2}; \frac{3}{4}\right).
\end{align}
It is of interest to note that it becomes evident that
$$ {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; \frac{3}{4}\right) = \frac{2 \, \pi}{3 \, \sqrt{3}}. $$
There are several recurrence relations for the ${}_{2}F_{1}$-hypergeometric functions for which relations can be developed. An example is to use:
\begin{align}
(b-c+1) \, {}_{2}F_{1}\left(a,b; c; x\right) &= (2b - c+2 +(a-b-1) \, x) \, {}_{2}F_{1}(a,b+1; c; x) \\
& \hspace{20mm}+ (b+1)(x-1) \cdot \, {}_{2}F_{1}(a,b+2; c; x)
 \end{align}
which leads to
\begin{align}
(n+3) \, a_{n+4} = (5 n +6) \, a_{n+2} - 4 n \, a_{n}. 
\end{align}
Since $a_{0} = \pi/3$ and $a_{1} = \ln(2+\sqrt{3})$ then the remaining terms can be developed.
