# 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.

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$$\int_0^1\log\frac{\cos\frac{x}{2}}{\cos\left(\frac{x}{2}+\frac\pi4\right)} dx\leq\log\,2$$ then? – J. M. Jul 15 '12 at 12:01
@J.M.: The inequality you noted is a magic one for me :) But I just know that for $0<x<y$, $\ln(\frac{y}{x})<\frac{y-x}{x}$ and this gives a big upper bound than $ln(2)$. – Babak S. Jul 15 '12 at 13:59
@Babak: huh? I just simplified that nasty square root within the logarithm, you know... – J. M. Jul 15 '12 at 14:01

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.
Well, you were using convexity, so Jensen's came to mind. $e^x$ is the most common convex function and your expression had $\ln$'s in it, so I tried to turn it into a Jensen problem using $e^x$. – Zarrax Jul 15 '12 at 14:23
By the way, using one trapezoid isn't enough as you'll get a result that is larger than $2\ln2$ (or $\ln2$ in the original question.) – Zarrax Jul 30 '12 at 22:30