Tight integral inequality We define the integral 
$$
J_k = \int_0^\pi (a+bx) \frac{\sin^3x}{1 + k \cos^2x}\,\mathrm{d}x\,,
$$
and
$$
J = J_1 = \int_0^\pi (a+bx) \frac{\sin^3x}{1 + \cos^2x}\,\mathrm{d}x\,,
$$
where $a,b \in \mathbb{R}$. 

Prove that $J_{1/k} / J$ is independent of $a,b$ and that $$ J_{1/3}
 >J \cdot (\log 3)\,,
$$
  without a calculator. 

Using the following identity
$$
\int_0^{\pi} x f(\sin x) \,\mathrm{d}x 
= \int_0^{\pi} f(\sin x)\,\mathrm{d}x
$$
I think was able to prove the first part of the question. However I am stuck on the second part, is there an easier way than explicitly calculating both integrals and comparing?
 A: I dont know any explicit simplification but the function is such that the integral for any k can be calculated so we first calculat the integral for general k. Let $J_k=T $ thus $T=a\int _0 ^{\pi} \frac{\sin^3 (x}{1+k\cos^2 (x)}+M $ where $M $ is the next part of the integral. Thus $M=b\int _0 ^{\pi} (\pi-x)\frac {sin^3 (x)}{1+k\cos^2 (x)} $ now let $\cos (x)=u $ thus $-sin (x)dx=du $ thus we have $(1+b)M=\int _{-1} ^1b\pi \frac {1-u^2}{1+ku^2} $ so we plug this value in original integral thus we have $T=(\frac {(1+b)a+b\pi}{(1+b)})\int _0 ^{\pi}\frac {1}{\frac {1}{k}+u^2}-(\frac {u^2+\frac {1}{k}}{\frac {1}{k}+u^2}-\frac {\frac {1}{k}}{\frac {1}{k}+u^2}) $ all this can be easily integrated and then substituting the limits the  final value of the  integral is $T=2\frac {(1+b)a+b\pi}{(1+b)k^2}(k+1)\sqrt {k}\arctan (\sqrt {k}) $ . Let $d=2\frac {a (1+b)+b\pi}{1+b} $ its common in all k's thus cancels .Now we put $k=1/3,1$ so $\frac {T _{\frac {1}{3}}}{T_1}=\frac {4}{\sqrt {3}}>1>\log (3) $
