The value of I=int_{sqrt{log _e 2}}^{sqrt{log | vedclass

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(A) Let $I=\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin (\log 6-x^2)} d x$.Put $x^2=t$,then $2x \, dx = dt$,so $x \, dx = \fra

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$\frac{1}{4} \log _e \frac{3}{2}$

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(A) Let $I=\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin (\log 6-x^2)} d x$.Put $x^2=t$,then $2x \, dx = dt$,so $x \, dx = \frac{1}{2} dt$.The limits change as follows: when $x = \sqrt{\log 2}$,$t = \log 2$; when $x = \sqrt{\log 3}$,$t = \log 3$.Thus,$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t}{\sin t + \sin (\log 6 - t)} dt \quad \dots (i)$Using the property $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$,we get:$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2 + \log 3 - t)}{\sin (\log 2 + \log 3 - t) + \sin (\log 6 - (\log 2 + \log 3 - t))} dt$$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin (\log 6 - t) + \sin t} dt \quad \dots (ii)$Adding $(i)$ and $(ii)$:$2I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6 - t)}{\sin t + \sin (\log 6 - t)} dt$$2I = \frac{1}{2} \int_{\log 2}^{\log 3} 1 \, dt = \frac{1}{2} [t]_{\log 2}^{\log 3} = \frac{1}{2} (\log 3 - \log 2) = \frac{1}{2} \log \left(\frac{3}{2}\right)$$I = \frac{1}{4} \log \left(\frac{3}{2}\right)$.

The value of $I=\int_{\sqrt{\log _e 2}}^{\sqrt{\log _e 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log _e 6-x^2\right)} d x$ is

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