Discussion Forum

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 

Answer:

Explanation:

  put $x^{2}=t \Rightarrow x dx= \frac{dt}{2}$

 $\therefore$ $I=\int_{\log 2}^{\log 3} \frac{ \sin t \frac{dt}{2}}{ \sin t+\sin ( log 6-t)}$ ...(i)

 Using $\int_{a}^{b} f(x) dx= \int_{a}^{b}  f(a+b-x)dx$

$I=\frac{1}{2}\int_{\log 3}^{\log 2} \frac{\sin(\log 2+\log 3-t)}{\sin(  \log 2+\log 3-t)+\sin (\log 6-(\log 2+\log 3-t)) }dt$

 $I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin(\log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$

$\therefore$   $I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin(\log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$......(i)

On adding Eqs.(i) and (ii)  we get

 $2l=I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin t+ \sin(log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$

 $\therefore$   $2I=\frac{1}{2}(t)_{\log2}^{\log 3}= \frac{1}{2} (\log 3-\log 2)$

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