Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$ 
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$

Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\,\mathrm dt$ 
How do you get this in the form of $t$ instead of $x$ using $\sin A=\dfrac{2t}{1+t^2}$ ?
$$=\int_0^{1/\sqrt3}\frac{2-2\sin^2\frac{1}{2}x}{1-\frac{2t}{1+t^2}}\mathrm dt\,??$$

The Weierstrass substitution sets $t = \tan \frac{x}{2}$ and it's possible that's what you mean to do. In that case, $$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2} \mathrm{sec}^2 \frac{x}{2} = \frac{1}{2} (1 + \tan^2 \frac{x}{2}) = \frac{1 + t^2}{2}$$ which gives you $$\mathrm{d}x = \frac{2}{1+t^2} \mathrm{d}t.$$
