Why does this proof fail? Catalan Constant's Exact Value. A failed attempt at finding an exact value for Catalan's Constant "C"
Definition :
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(x)}}{\sqrt{1+\cos{(x)}^2}} = \pi \ln(1+\sqrt{2})
 $$
Let
$$I(b) = \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(bx)}}{\sqrt{1+\cos{(x)}^2}}$$
such that $ I(b=1) $ is our desired integral.
Integrate both sides with respect to b (Ad = antiderivative)
$$ I^{Ad} (b)= -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} $$
I shall restrict b to be an odd integer greater than $0$
I Examine the values given by the integral :
$$ p = 1 \to \left (\frac{\pi}{4}\right)$$
$$ p = 3 \to \left(2-\frac{3 \pi}{4}\right) $$
$$ p = 5  \to \left(\frac{13 \pi }{4} - 10\right) $$
I divide these values into Pi Numbers ( those with $\pi$) and Rational Numbers (those without it).
The Pi Numbers will follow a known sequence called The Central Delannoy Numbers $D(n)$
$$ D(n) = \sum_{k=0}^{n} \binom{n}{k}^{2} 2^k $$
Putting this into b terms .
$$ \Pi (b) = \frac{\pi}{4}  \sum_{k=0}^{\frac{b-1}{2}}\binom{\frac{b-1}{2}}{k}^2\, 2^k $$
For the Rational Numbers $R(b)$ we notice that since b is an integer then the $\cos(bx)$ may be expanded with Chebyschev Polynomials in the form :
$$ \cos(bx) = b \sum_{k=0}^{b} (-2)^{k} \frac{(b+k-1)!}{(b+k)!(2k)!} (1-\cos(x))^{k} $$
Since
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos(x)^{b}}{\sqrt{1+\cos(x)^2}} = \frac{\sqrt{\pi}}{4} \frac{\Gamma(\frac{b}{4}+\frac{1}{4})}{\Gamma(\frac{b}{4}+\frac{3}{4})}$$
for odd integers greater than $0$
We can give $R(b)$ as
$$ \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}} $$
Together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} = \sqrt{2}  \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right)\left(\frac{\pi}{4}  \sum_{k=0}^{\frac{b-1}{4}}\binom{\frac{b-1}{2}}{k}^2\, 2^k - \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}}\right)$$
All that is left is to take the derivative with respect to b and let $b = 1$ ( I am unsure of the constant that is generated when we integrate with respect to b  , don't know if it goes to $0$ after we take the derivative again )
I do not know how to take the derivative of b in its current form so i decided to change it into integral form.
The Central Delannoy Numbers can be given with :
$$ D(n) = \frac{1}{\pi} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{1}{\sqrt{-t^2+6t-1}} \frac{1}{t^{n+1}} \,dt $$
The generating function for $R(b) $ is
$$ \frac{\sin^{-1}\left(\frac{4x}{(1-x)^2}\right)}{2\sqrt{x^2-6x+1}} $$
Putting it all together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}}= \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right) \left(\frac{1}{2 \sqrt{2}} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}} \, dt -\sqrt{2} \,\lim_ {t \to 1} \int \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}}\, dt \right) $$
Unfortunately, taking the derivative with respect to b and letting $b=1$ yields
$$ \frac{1}{2} \lim_{t \to 1} \int \frac{\ln{t}}{t} \frac{1}{\sqrt{-t^2+6t-1}} \,dt $$
which cancels the Catalan Constant on the LHS.
Question(s) : Does anybody see another way to proceed which could possibly lead to better results?
Does anyone see any mistakes?
Thank you kindly for your help and time.

Integrate by parts
\begin{align}
\int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2{x}}} dx
=-
 \int_{\frac{\pi}{2}}^{\pi} x d(\sinh^{-1}\cos{x})
=\pi \ln(1+\sqrt{2}) - J(1)
\end{align}
where $J(a)= \int^{\frac{\pi}{2}}_{0} \sinh^{-1}(a\cos t) dt$. Evaluate
$$
J’(a)= \int^{\frac{\pi}{2}}_{0} \frac{\cos{t}}{\sqrt{1+a^2\cos^2{t}}} dt
=\frac1a \sin^{-1}\frac a{\sqrt{1+a^2}}= \frac{\tan^{-1}a}a
$$
$$J(1)= \int_0^1 \frac{\tan^{-1}a}ada=-\int_0^1\frac{\ln a}{1+a^2}da=C
$$
Thus
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2x}} = \pi \ln(1+\sqrt{2})
 $$
