Is there a closed form for $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x$? When I encountered the integral $$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x,$$
where n is a non-negative integer, I, as usual, used the substitution $x\mapsto \frac{\pi}{2}-x$ to transforms
$$I=
\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(\sin x+\cos x)^{5}} d x
$$
Then $$
\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{4}} \\
&=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\cos ^{4}\left(\frac{\pi}{4}-x\right)} \\
&=\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d(\tan x)\\&= \frac{1}{3}\\ \therefore I&= \frac{1}{6}
\end{aligned}
$$
Similarly, we can generalise $I$ to
$$I_n =\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(\sin x+\cos x)^{2 n+1}} $$
$$
\begin{aligned}
2 I_{n} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin x+\cos x}{(\sin x+\cos x)^{2 n+1}} d x \\
&=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left[\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)\right]^{2 n}}\\&=\frac{1}{2^{n}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2 n} x d x
\end{aligned}
$$
Letting $t=\tan x$ yields
$$
\begin{aligned}
I_{n} &=\frac{1}{2^{n}} \int_{0}^{1}\left(1+t^{2}\right)^{n-1} d t \\
&=\frac{1}{2^{n}} \sum_{k=0}^{n-1}\left(\begin{array}{l}
n-1 \\
\quad  k
\end{array}\right) \int_{0}^{1} t^{2 k} d t \\
&=\frac{1}{2^{n}} \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \frac{1}{2 k+1}
\end{aligned}
$$
My question is how to simplify the last sum?

Something which could be interesting
$$I_n=\frac 1 {2^n} \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{1}{2 k+1}=2^{-n} \, \,_2F_1\left(\frac{1}{2},1-n;\frac{3}{2};-1\right)$$ generates the unknown (?) sequence
$$\left\{\frac{1}{2},\frac{1}{3},\frac{7}{30},\frac{6}{35},\frac{83}{630},\frac{73}{6
   93},\frac{523}{6006},\frac{476}{6435},\frac{14051}{218790},\frac{13103}{230945},
   \frac{32867}{646646},\frac{31130}{676039},\cdots\right\}$$ but
$$\lim_{n\to \infty } \,( n\,I_n)=\frac 12$$
Let $n=10^k$ and compute
$$\left(
\begin{array}{cc}
k & 10^k\, I_{10^k} \\
 1 & 0.5673645240 \\
 2 & 0.5051036873 \\
 3 & 0.5005010035\\
 4 & 0.5000500100 \\
 5 & 0.5000050001 \\
 6 & 0.5000005000
\end{array}
\right)$$
