Mathematics - Definite Integration Question with Solution | TestHub

Put $x^2=t\Rightarrow xdx=\frac{dt}{2}$ $\therefore I={\int }_{\log 2}^{\log 3}\frac{\sin t\cdot \frac{dt}{2}}{\sin t+\sin (\log 6-t)}$ Using, ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$ $$\begin{array}{rl}& I=\frac{1}{2}{\int }_{\log 2}^{\log 3}\frac{\sin (\log 2+\log 3-t)}{\sin (\log 2+\log 3-t)+\sin }dt\\ & I=\frac{1}{2}{\int }_{\log 2}^{\log 3}\frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin (t)}dt\\ & \therefore I={\int }_{\log 2}^{\log 3}\frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t}dt\dots (\mathrm{li})\end{array}$$ On adding Eqs. (i) and (ii), we get $$2I=\frac{1}{2}{\int }_{\log 2}^{\log 3}\frac{\sin t+\sin (\log 6-t)}{\sin (\log 6-t)+\sin t}dt$$ $$\begin{array}{c}\therefore 2I=\frac{1}{2}(t{)}_{\log 2}^{\log 3}=\frac{1}{2}(\log 3-\log 2)\\ \Rightarrow I=\frac{1}{4}\log \left(\frac{3}{2}\right)\end{array}$$