Question

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

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

Answer 1

Answer: (A)

$\frac{1}{4}\ln \frac{3}{2}$

Let $I={\int }_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}\frac{x\sin x^2}{\sin x^2+\sin (\ln 6-x^2)}dx$

Now, substitute $ln6-x^2=t$

Thus, when $x=\sqrt{\ln 3}\to ,t=\ln 6-\ln 3=\ln 2$

Also, when $x=\sqrt{\ln 2}\to ,t=\ln 6-\ln 2=\ln 3$

$\Rightarrow -2xdx=dt$
Thus, $\Rightarrow dx=\frac{-dt}{2\sqrt{\ln 6-t}}$

Substituting the above values in $I$, we get

$I={\int }_{\ln 3}^{\ln 2}\frac{\sqrt{\ln 6-t}\sin (\ln 6-t)}{\sin t+\sin (\ln 6-t)}\frac{-dt}{2\sqrt{\ln 6-t}}$

$I=\frac{-1}{2}{\int }_{\ln 3}^{\ln 2}\frac{\sin (\ln 6-t)}{\sin (\ln 6-t)+\sin t}dt$

$I=\frac{1}{2}{\int }_{\ln 2}^{\ln 3}\frac{\sin (\ln 6-t)}{\sin (\ln 6-t)+\sin t}dt$...............(1)

Now also, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Thus, expression (1) can also be written as,

$I=\frac{1}{2}{\int }_{\ln 2}^{\ln 3}\frac{\sin (t)}{\sin \left(t\right)+\sin (\ln 6-t)}dt$...............(2)

Adding equations (1) and (2), we get,

$\Rightarrow 2I=\frac{1}{2}{\int }_{\ln 2}^{\ln 3}1dt$
$\Rightarrow I=\frac{1}{4}\left[\ln 3-\ln 2\right]=\frac{1}{4}\ln \frac{3}{2}$
Hence, option 'A' is correct.