Show that $$\frac 12+\sum_{n=0}^\infty\frac{(-1)^n}{(5+6n)(7+6n)}=\frac{\pi}6$$
My proof uses the Leibniz series, $$1-\frac 13+\frac 15-\frac 17+\frac 19-\frac 1{11}+\cdots = \frac{\pi}4$$
\begin{align*}&\qquad 1-\frac 13+\frac 15-\frac 17+\frac 19-\frac 1{11}+\cdots \\ &= 1+\bigg(\frac 15-\frac 17\bigg)-\bigg(\frac 1{11}-\frac 1{13}\bigg)+\cdots -\bigg(\frac 13-\frac 19+\frac 1{15}-\frac 1{21}+\cdots\bigg) \\ &= 1+2\sum_{n=0}^\infty\frac {(-1)^n}{(5+6n)(7+6n)}-\frac{\pi}{12}\end{align*} $$\therefore \frac 12+\sum_{n=0}^\infty\frac {(-1)^n}{(5+6n)(7+6n)}=\frac{\pi}6$$
My question is, how would one solve the problem without the Leibniz series? If one didn't know of the series, I'm not asking for a way to solve the problem through deriving the series first, I'm asking for a way to solve the problem without using the series at all.
I am curious because using the series the proof is incredibly short, clever and elegant. Thus, to me, it begs the question on how to solve the problem without the series, whilst still keeping clever and elegant (concision aside). I have no idea, hence why I have posted a question here. I assume some creativity is perhaps involved, vaguely speaking I suppose. Any thoughts?
Thank you in advance.
$$\frac{1}{(6n+5)(6n+7)}=\frac12\left(\frac1{6n+5}-\frac1{6n+7}\right) =\frac12\int_0^1(t^{6n+4}-t^{6n+6})\,dt=\frac12\int_0^1t^{6n+4}(1-t^2)\,dt$$ So $$\sum_{n=0}^\infty\frac{(-1)^n}{(6n+5)(6n+7)} =\frac12\int_0^1\sum_{n=0}^\infty(-1)^nt^{6n+4}(1-t^2)\,dt =\frac12\int_0^1\frac{t^4(1-t^2)}{1+t^6}\,dt$$ and now you have a nice integral to do.