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!
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1$\begingroup$ See also: math.stackexchange.com/questions/439851/… and math.stackexchange.com/questions/578957/… and math.stackexchange.com/questions/2074951/… $\endgroup$– lab bhattacharjeeCommented Jan 15, 2017 at 5:58
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1$\begingroup$ The basic idea: math.stackexchange.com/questions/2075867/an-integral-with-2017/… $\endgroup$– lab bhattacharjeeCommented Jan 15, 2017 at 5:58
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1 Answer
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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.
answered Jan 15, 2017 at 3:49