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. :(
$\begingroup$ $\endgroup$
1 
- $\begingroup$ first substitute x^2 = t and then write the denominator as the product form then another trivial substitution will make your job get done $\endgroup$user14618– user146182011-08-15 12:46:37 +00:00Commented Aug 15, 2011 at 12:46
Add a comment |
1 Answer
$\begingroup$ $\endgroup$
9 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 $\textbf{cancelled out. }$
- $\begingroup$ Thanks, that does the job. I was trying to use $f(a-x)=f(x)$ property of the definite integral, and that surely doesn't seem to work :( $\endgroup$user– user2011-08-15 10:28:46 +00:00Commented Aug 15, 2011 at 10:28
- $\begingroup$ @Chandrasekhar: Hi, how will the second formula aid in calculating the integral? After I apply it, I get: $I = \frac12 \int_{\ln 2}^{\ln 3} \frac{\sin(\ln6 - t)}{\sin(\ln6 - t) + \sin(t)} \ dt$ $\endgroup$jrand– jrand2011-10-09 04:59:00 +00:00Commented Oct 9, 2011 at 4:59
- $\begingroup$ @jrand: Then you will simply have $\int_{\ln{2}}^{\ln{3}} dt$ once the numerator and denominator get's cancelled out. $\endgroup$user9413– user94132011-10-11 08:10:51 +00:00Commented Oct 11, 2011 at 8:10
- $\begingroup$ @Chandrasekhar: Hi, are you using a trigonometric identity to simplify $\frac{\sin(\ln6 - t)}{\sin(\ln6 - t) + \sin(t)}$? Because I am looking through the tables, I see one which handles the sum of $\sin$ in the denominator, but that ends up giving me cosine terms. $\endgroup$jrand– jrand2011-10-11 14:34:26 +00:00Commented Oct 11, 2011 at 14:34
- 1$\begingroup$ @Chandrasekhar: Hi. Sorry, the final answer should be $\frac{1}{4} \int_{\ln{2}}^{\ln{3}} dt$. The reason is $I = \frac12 \int \frac{\sin(A)}{\sin(A) + \sin(B)} \ dt= \frac12 \int \frac{\sin(B)}{\sin(A) + \sin(B)} \ dt$. Then, $2I = \frac12 \int \ dt$. Then, $I = \frac14 \int \ dt$. The MATLAB code to check this can be found here: pastebin.com/RXgNAjkK . Thank you for the technique. $\endgroup$jrand– jrand2011-10-16 14:51:51 +00:00Commented Oct 16, 2011 at 14:51