I haven't seen it before-1

$\large \displaystyle\sum_{r=1}^{\infty}(-1)^{r+1}\sin^r x$ represents a sum of infinite GP (since $|\sin x|<1$ ) with first term $\sin x$ and common ratio $-\sin x$ and equals $\dfrac{\sin x}{1+\sin x}$ . Thus:

$\large\mathfrak T=\displaystyle\int_0^{\pi/6}\dfrac{\cos x\sin x\mathrm{d}x}{1+\sin x}$ Put $\sin x=t$ such that $\cos x\mathrm{d}x=\mathrm{d}t$ .

$\large \mathfrak T=\displaystyle\int_0^{1/2}\dfrac{t\mathrm{d}t}{1+t}$

$\large =\displaystyle\int_0^{1/2}\left(1-\dfrac{1}{1+t}\right)\mathrm{d}t$

$\large =\left|t-\ln(1+t)\right|_0^{1/2}=1/2-\ln(3/2)$

$\Large =\ln\left(\dfrac{2\sqrt e}{3}\right)$

$\Huge \therefore~2\times 2\times 3=\boxed{12}$