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How to integrate: $\int\limits_{\sqrt{\ln{2}}}^{\sqrt{\ln{3}}} \frac{ x \cdot \sin(x^{2})}{\sin(x^{2}) + \sin(\ln{6}-x^{2})} \ \mathrm dx$

Any idea of how to solve. Tried using substitution, $x^2=t$ but didn't succeed. :(

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    first substitute x^2 = t and then write the denominator as the product form then another trivial substitution will make your job get done2011-08-15

1 Answers 1

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Here are some hints:

  • Substituting $x^{2}=t$ as you did is correct. After doing this your integral becomes $I =\frac{1}{2} \cdot \int\limits_{\ln{2}}^{\ln{3}} \frac{ \sin{t}}{\sin{t} + \sin(\:\ln{6}-t)} \ dt \qquad \cdots (1)$

  • Next, use this formula: $\int\limits_{a}^{b} f(x) \ dx = \int\limits_{a}^{b} f(a+b-x) \ dx$

Also you have $ I = \frac{1}{2}\int\limits_{\ln{2}}^{\ln{3}} \frac{\sin(\ln{6}-t)}{\sin{t} + \sin(\ln{6}-t)} \ dt \qquad \cdots (2) $

Add $(1) + (2)$.

Now $\text{you may ask why I am doing this}$: Because $\ln{2} + \ln{3} = \ln{6}$ which is in the denominator, and gets $\textbf{cancelled out. }$

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    @jrand: No problem. Glad i could help.2011-10-16