Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form. 
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos x }dx $$ 
Substituting $t=\tan(\frac{x}{2})$,
$$\int_{\tan(\frac{\pi}{12})}^{\tan(\frac{\pi}{6})} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
$$2\int_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}} \dfrac{2t(1-t^2)}{(1+t^2)^3}dt$$
Substituting $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$ Could somebody please tell me where I have gone wrong? Also could someone please tell me how to change the limits of the definite integral throughout? 

Integrate directly as follows
\begin{align}
\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } 
& =\frac12 \int _{ \pi /6 }^{ \pi /3 } \frac{(\sin x +\cos x)^2-1 }{ \sin x+\cos x }dx \\
&= \frac{1}2\int _{ \pi /6 }^{ \pi /3 } \left(\sqrt2\cos(\frac\pi4-x)
-\frac1{\sqrt2\cos(\frac\pi4-x)}\right)dx\\
&= \frac12\left[{\sqrt2}\sin(\frac\pi4-x)dx
-\frac1{\sqrt2} \ln \cot(\frac\pi8-\frac x2)\right]_{\pi/6}^{\pi/3}\\
&= \frac{\sqrt3-1}2-\frac1{2\sqrt2}\ln \frac{\cot\frac{5\pi}{24}}{\cot\frac{7\pi}{24}}
\end{align}
