Question

The value of ${\int }_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}\frac{x\sin x^2}{\sin x^2+\sin \left(\ln 6-x^2\right)}dx$ is

• A. $\frac{1}{4}\ln \frac{3}{2}$
• B. $\frac{1}{2}\ln \frac{3}{2}$
• C. $\ln \frac{3}{2}$
• D. $\frac{1}{6}\ell n\frac{3}{2}$

Answer 1

Answer: (A)

$<br/>I=\frac{1}{2}{\int }_{\sqrt{\ln 2}}^{\sqrt{\ell n3}}\frac{2x\sin x^2}{\sin x^2+\sin \left(\ln 6-x^2\right)}dx<br/>$
Let $x^2=t\Rightarrow 2xdx=dt$
Also, when $x=\sqrt{\ell n2},t=\ell n2$
when $x=\sqrt{\ln 3},t=\ell n3$
$<br/>\therefore I=\frac{1}{2}{\int }_{\ell n2}^{\ell n3}\frac{\sin tdt}{\sin t+\sin (\ell n6-t)}\dots (1)<br/>$
$<br/>I=\frac{1}{2}{\int }_{\ln 2}^{\ln 3}\frac{\sin (\ell n6-t)}{\sin t+\sin (\ln 6-t)}dt\dots (2)<br/>$
Adding values of $I$ in equations (1) and (2)
$<br/>2I=\frac{1}{2}{\int }_{\ell n2}^{\ln 3}1dt=\frac{1}{2}(\ln 3-\ln 2)=\frac{1}{2}\ell n\frac{3}{2}\Rightarrow I=\frac{1}{4}\ell n\frac{3}{2}<br/>$