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How to evaluate $$I=\int_{\sqrt{\ln2}}^{\sqrt{\ln3}}\frac{x\sin\left ( x^{2} \right )}{\sin\left ( x^{2} \right )+\sin\left ( \ln6-x^{2} \right )}\,\mathrm dx$$ I tried to substitute $x^2=t$ so $$I=\frac{1}{2}\int_{\ln2}^{\ln3}\frac{\sin t}{\sin t+\sin\left ( \ln6-t \right )}\, \mathrm{d}t $$ but I got stuck.Any hint?Thx!

Renascence_5.
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    See also: http://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 and http://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3 and http://math.stackexchange.com/questions/2074951/find-the-value-of-int1-1-x-ln1x-2x-3x-6x-dx – lab bhattacharjee Jan 15 '17 at 05:58
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    The basic idea: http://math.stackexchange.com/questions/2075867/an-integral-with-2017/2075921#2075921 – lab bhattacharjee Jan 15 '17 at 05:58

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Use the fact $\displaystyle\int_{a}^{b}f\left ( x \right )\mathrm{d}x=\int_{a}^{b}f\left ( a+b-x \right )\mathrm{d}x$ here we get $$I=\frac{1}{2}\int_{\ln2}^{\ln3}\frac{\sin \left ( \ln6-t \right )}{\sin t+\sin\left ( \ln6-t \right )}\, \mathrm{d}t$$ hence $$2I=\frac{1}{2}\int_{\ln2}^{\ln3}\, \mathrm{d}t$$ and the answer will follow.

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Renascence_5.
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