An Integral Inequality Problem How to establish the Integral Inequalities : $$ \displaystyle \int_0^1 \ln \sqrt{\dfrac{1-\cos x}{1+\sin x}} \,dx < \dfrac{1}{2}\ln 2$$
My attepmt :
We have $\displaystyle $$(ii) \displaystyle \int_0^1 \ln \sqrt{\dfrac{1-\cos x}{1+\sin x}} \,dx = \int_0^1 \ln \dfrac{\sin(x/2)}{\sin (\frac{x}{2}+\frac{\pi}{4})} = \int_0^1 \ln \sin(x/2) - \ln \sin (\frac{x}{2}+\frac{\pi}{4}) \,dx$
Now, using $I= \sum\limits_{n=1}^{\infty} \frac{\cos nx}{n} = −\ln(2 \sin x/2)$, and after a bit of calculation I could reduce the above integral into the series: $I=\sum\limits_{n=1}^{\infty} \dfrac{(−1)^{n+1}cos(2n+1)−2sin(2n+1)−1}{(2n+1)^2}$
So, now it suffices to show that $I < \frac{1}{2}\ln 2$. But I'm stuck at this point . \ln 2$$
Any ideas or suggestions ? Any ideas on a different way that can bypass this calculation altogether ?
Thank you :)
 A: This is an improper integral, but because for $x \approx 0$ we have
$$
\ln\sqrt\frac{1-\cos x}{1+\sin x}\approx\ln\left(\frac{x}{\sqrt 2}\right)
$$
the integral exists.
So we can procede and we have
$$
1-\cos x <2\\
1+\sin x \geq 1
$$
for $x\in[0,1]$, so
$$
\frac{1-\cos x}{1+\sin x}<2
$$
But you can do better, given that $1-\cos x\leq1$, so you get $I\leq(1/2)\log1=0$. The value of the integral is in fact $I\approx-1.542$.
A: In the same spirit as Enzotib's answer, $$f(x)=\ln \sqrt{\dfrac{1-\cos x}{1+\sin x}} $$ is an increasing function in the range of $x$ considered for its integration and $$f(x) \le \frac{1}{2} \log \left(\frac{1-\cos (1)}{1+\sin (1)}\right) \simeq -0.693875$$ Then $$\displaystyle \int_0^1 \ln \sqrt{\dfrac{1-\cos x}{1+\sin x}} \,dx <-0.693875 \lt \dfrac{1}{2}\ln 2 $$
A: Hint: $\dfrac12\ln2=\ln\sqrt2=\displaystyle\int_0^1\ln\sqrt2~dx$, so proving that $\dfrac{1-\cos x}{1+\sin x}<2$ on $(0,1)$ should suffice, 
which can easily be shown using the fact that $\dfrac{1-\cos x}2=\sin^2\dfrac x2$ , and $\sin x=2\sin\dfrac x2\cos\dfrac x2$ .
A: You have $$\displaystyle \int_0^1 \ln \sqrt{\dfrac{1-\cos x}{1+\sin x}} \,dx$$ =
$$\dfrac{1}{2}\int_0^1 \left(\ln(1-\cos x)-\ln(1+\sin x)\right)dx$$=
$$\dfrac{1}{2}\int_0^1\left[\ln2+2\ln \left(\sin \dfrac{x}{2}\right)-2\ln\left(\sin\dfrac{x}{2}+\sin\left(\dfrac{x}{2}+\dfrac{\pi}{2}\right) \right) \right]\lt \dfrac{1}{2}\ln2+\int_0^10 dx $$
since $\sin x$ increases in $(0,1)$, $\sin \left(\dfrac{x+\pi}{2}\right) \gt 0 \implies \ln\left(\sin\dfrac{x}{2}+\sin\dfrac{x+\pi}{2}\right) \gt \ln\left(\sin \dfrac{x}{2}\right) $
