Integration (5)

Using the identity $\sin^{3}(t) = \frac{3\sin(t) - \sin(3t)}{4}$ , we can perform integration-by-parts according to:

$\int_{0}^{\pi} t(\frac{3\sin(t) - \sin(3t)}{4}) dt;$

or $\frac{3}{4} \int_{0}^{\pi} t \sin(t) dt - \frac{1}{4}\int_{0}^{\pi} t \sin(3t) dt;$

or $\frac{3}{4}[-t \cos(t) + \int \cos(t) dt] + \frac{1}{4}[\frac{t \cos(3t)}{3} - \int \frac{\cos(3t)}{3}] dt$ ;

or $\frac{3}{4}[-t \cos(t) + \sin(t)] + \frac{1}{4}[\frac{t \cos(3t)}{3} - \frac{\sin(3t)}{9}]|_{0}^{\pi};$

or $\frac{3\pi}{4} - \frac{\pi}{12} = \frac{2\pi}{3} \Rightarrow A = 2, B = 3$ ;

or $\frac{B}{A} = \boxed{\frac{3}{2}}.$