Question
The value of $$\int\limits_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {\frac{{x\,\sin \,{x^2}}}{{\sin \,{x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}dx} ,$$ is-
A.
$$\frac{1}{4}\ell n\frac{3}{2}$$
B.
$$\frac{1}{2}\ell n\frac{3}{2}$$
C.
$$\ell n\frac{3}{2}$$
D.
$$\frac{1}{6}\ell n\frac{3}{2}$$
Answer :
$$\frac{1}{4}\ell n\frac{3}{2}$$
Solution :
$$\eqalign{
& I = \frac{1}{2}\int_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {\frac{{2x\,\sin \,{x^2}}}{{\sin \,{x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}dx} \cr
& {\text{Let }}\,{x^2} = t\,\,\, \Rightarrow 2x\,dx = dt \cr
& {\text{Also when}}\,x = \sqrt {\ell n2} ,\,\,t = \ell n2 \cr
& {\text{when }}x = \sqrt {\ell n3} ,\,\,t = \ell n3 \cr
& \therefore I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {\frac{{\sin \,t\,dt}}{{\sin \,t + \sin \left( {\ell n6 - t} \right)\,}}.....(1)} \cr
& {\text{Using }}\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} \cr
& {\text{We get, }}I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {\frac{{\sin \,\left( {\ell n6 - t} \right)}}{{\sin \,t + \sin \left( {\ell n6 - t} \right)\,}}dt.....(2)} \cr} $$
Adding values of $$I$$ in equation (1) and (2)
$$\eqalign{
& 2I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {1\,dt} = \frac{1}{2}\left( {\ell n3 - \ell n2} \right) = \frac{1}{2}\ell n\frac{3}{2} \cr
& \Rightarrow I = \frac{1}{4}\ell n\frac{3}{2} \cr} $$