JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 10)
The value of $$\,\int\limits_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {{{x\sin {x^2}} \over {\sin {x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}\,dx} $$ is
$${1 \over 4}\,\ell n{3 \over 2}$$
$$\,{1 \over 2}\,\ell n{3 \over 2}$$
$$\ell n{3 \over 2}$$
$$\,\,{1 \over 6}\,\ell n{3 \over 2}$$
Explanation
$${x^2} = t \Rightarrow 2x\,dx = dt$$
$$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt} $$ and $$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt} $$
$$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}} $$.
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