A Definite Problem

Relevant wiki: Integration Tricks

Substitute $x^2=t$ such that $2x\mathrm{d}x=\mathrm{d}t$

$\color{#3D99F6}{\mathcal I}=\dfrac 12\int_{\ln 2}^{\ln 3}\dfrac{\sin t~\mathrm{d}t}{\sin t+\sin (\ln 6-t)}$

Now use $\small{\color{teal}{\displaystyle\int_a^bf(x)\mathrm{d}x=\displaystyle\int_a^b f(a+b-x)\mathrm{d}x}}$ so that:

$\color{#3D99F6}{\mathcal I}=\dfrac 12\int_{\ln 2}^{\ln 3}\dfrac{\sin (\ln 6-t)~\mathrm{d}t}{\sin t+\sin (\ln 6-t)}$

Adding we get $\mathcal I=\dfrac 14\int_{\ln 2}^{\ln 3}\mathrm{d}t=\dfrac 14\ln \left(\dfrac 32\right)\approx \boxed{0.101}$

I did it the long way, but that's alright, someone ought to show the long way too :p