Integration of a trigonometric function

Question

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

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

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

score 16 · Accepted Answer · 2011-10-11 14:40:01Z

up vote 16 down vote accepted

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 $\textbf{cancelled out. }$

edited Oct 11 '11 at 14:40

answered Aug 15 '11 at 10:20

user9413

	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 :( – user Aug 15 '11 at 10:28
	@Chandrasekhar: Hi, how will the second formula aid in calculating the integral? After I apply it, I get: $I = \frac12 \int_{\ln 2}^{\ln 3} \frac{\sin(\ln6 - t)}{\sin(\ln6 - t) + \sin(t)} \ dt$ – jrand Oct 9 '11 at 4:59
	@jrand: Then you will simply have $\int_{\ln{2}}^{\ln{3}} dt$ once the numerator and denominator get's cancelled out. – user9413 Oct 11 '11 at 8:10
	@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. – jrand Oct 11 '11 at 14:34
	@jrand: I have edited the answer. Hope that will help. – user9413 Oct 11 '11 at 14:40

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