The value of int_{sqrt{ln 2}}^{sqrt{ln 3}} fr | vedclass

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(A) Let $I = \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin (\ln 6 - x^2)} dx$.Substitute $x^2 = t$,so $2x dx = dt$ or $x dx = \

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

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(A) Let $I = \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin (\ln 6 - x^2)} dx$.Substitute $x^2 = t$,so $2x dx = dt$ or $x dx = \frac{1}{2} dt$.When $x = \sqrt{\ln 2}$,$t = \ln 2$. When $x = \sqrt{\ln 3}$,$t = \ln 3$.The integral becomes $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin (\ln 6 - t)} dt$.Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,we have $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 2 + \ln 3 - t)}{\sin (\ln 2 + \ln 3 - t) + \sin (\ln 6 - (\ln 2 + \ln 3 - t))} dt$.Since $\ln 2 + \ln 3 = \ln 6$,this simplifies to $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 6 - t)}{\sin (\ln 6 - t) + \sin t} dt$.Adding the two expressions for $I$:$2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t + \sin (\ln 6 - t)}{\sin t + \sin (\ln 6 - t)} dt = \frac{1}{2} \int_{\ln 2}^{\ln 3} 1 dt$.$2I = \frac{1}{2} [t]_{\ln 2}^{\ln 3} = \frac{1}{2} (\ln 3 - \ln 2) = \frac{1}{2} \ln \frac{3}{2}$.Therefore,$I = \frac{1}{4} \ln \frac{3}{2}$.

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

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