Integration of a trigonometric function

calculus integration

Solution 1:

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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