<s>[INST] prove Inequalities for integrals prove
$\frac{\pi}{6}+\frac{1}{3}\leq \int_0^\frac{\pi}{2}\frac{1+\cos(x)}{2+\sin(x)}dx \leq \frac{\pi}{4}+\frac{1}{2}$
I got to the point where  $\frac{1}{3} \leq f(x) \leq 1$,
so  $\frac{\pi}{6} \leq \int_0^\frac{\pi}{2}f(x)dx \leq \frac{\pi}{2}$ and where do I go from here , thanks 
 [/INST] The upper bound is easy:$$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x\leq \int_2^{\frac{\pi}{2}}\frac{1}{2}(1+\cos x)\mathrm{d}x=\pi/4+1/2.$$
For the lower bound, $$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{2+\sin x}\mathrm{d}x+\int_0^{\frac{\pi}{2}}\frac{\cos x}{2+\sin x}\mathrm{d}x.$$ The first term on the right hand side is $\geq \pi/6$. The second term $$\int_0^{\frac{\pi}{2}}\frac{\cos x}{2+\sin x}\mathrm{d}x=\int_0^1 \frac{1}{2+t}\mathrm{d}t=\ln 3-\ln 2 \geq 1/3.$$ 
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