Prove that: $ \int_{0}^{1} \ln \sqrt{\frac{1+\cos x}{1-\sin x}}\le \ln 2$ I plan to prove the following integral inequality:
$$ \int_{0}^{1} \ln \sqrt{\frac{1+\cos x}{1-\sin x}}\le \ln 2$$
Since we have to deal with a convex function on this interval i thought of considering the area of the trapeze that can be formed if we unify the points $(0, f(0))$ and $(1, f(1))$, where the function $f(x) =\ln \sqrt{\frac{1+\cos x}{1-\sin x}}$,  but things are ugly even if the method itself isn't
complicated. So,
I'm looking for something better if possible.
 A: Here's a (hopefully) corrected proof that uses convexity along with the trapezoid rule: You can rewrite what you're trying to prove as
$$ \int_{0}^{1} \ln {\frac{1+\cos x}{1-\sin x}}\,dx\le 2\ln 2$$
Let $f(x) = \ln {\frac{1+\cos x}{1-\sin x}} = \ln(1 + \cos x) - \ln (1 - \sin x)$.
Then 
$$f'(x) = -\frac{\sin x}{1 + \cos x} + \frac{\cos x}{1 - \sin x}$$
Using the tangent half-angle formula, this is the same as
$$-\tan(x/2) + \tan(x/2 + \pi/4)$$
Therefore
$$f''(x) = -(1/2)\sec^2(x/2) + 1/2\sec^2(x/2 + \pi/4)$$
Since $\sec$ is increasing on $(0,1/2 + \pi/4)$, we see that $f''(x) > 0$. So the integrand is convex. When applied to a convex function, the trapezoid rule always gives a result larger than the integral. But already with $2$ pieces, the trapezoid rule here gives
$$1/4(\ln(1 + \cos(0)) - \ln(1 - \sin(0)) + 2(\ln(1 + \cos(1/2)) - \ln(1 - \sin(1/2)))$$
$$ +\ln(1 + \cos(1)) - \ln(1 - \sin(1)) )$$
$$= 1.3831395912690787...$$
This is slightly less than $2\ln2 = 1.3862943611198906...$, so the original integral is less than $\ln 2$ as needed.
