Question

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

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

Answer 1

The correct option is A $\frac{1}{4}\log \frac{3}{2}$

$I\stackrel{\sqrt{log3}}{\underset{\sqrt{\log 2}}{\int }}\frac{x\sin x^2}{\sin x^2+\sin (\log 6-x^{)}}dx$

Let $x^2=t,xdn=\frac{dt}{2}$

$x=\sqrt{log3}\Rightarrow t=log3$

$x=\sqrt{log2}\Rightarrow t=log2$

$\Rightarrow I=\frac{1}{2}\stackrel{\sqrt{log3}}{\underset{\sqrt{\log 2}}{\int }}\frac{\sin t}{\sin t+\sin (log6-t)}dt$ -------(1)

use property

$[\therefore {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right]$

$I=\frac{1}{2}\stackrel{\sqrt{log3}}{\underset{\sqrt{\log 2}}{\int }}\frac{\sin (log6-t)}{\sin t+\sin (log6-t)}dt$ ------(2)

$\Rightarrow equation\left(1\right)+equation\left(2\right)$

$2I=\frac{1}{2}\stackrel{\sqrt{log3}}{\underset{\sqrt{\log 2}}{\int }}\frac{\sin (log6-t)+sint}{\sin (log6-t)+\sin t}$

$I=\frac{1}{4}\stackrel{\sqrt{log3}}{\underset{\sqrt{\log 2}}{\int }}dt=\frac{1}{4}(log3-log2)$

$I=\frac{1}{4}log\frac{3}{2}$