calculus - Cannot understand an Integral

$$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } =\frac { \sqrt { a } -b }{ 2 } +\frac { \sqrt { c } }{ 2 } \log(\sqrt { d } +\sqrt { e } -\sqrt { f } -g)$$
I had to solve the integral and get it in this form. However, I cannot understand what to do.
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 $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$
After integrating this last step, I had thought I would substitute back $u=1+t^2=1+\tan^2(\dfrac{x}{2})$. However, this doesn't seem to give me the required value. 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? This is the first time I have used a Weierstrass Substitution and so I don't know how to change the limits.
Please do help me!

Really, really grateful for your help!

Answer

Here is an alternate method you could use:

Multiply $\displaystyle\int\frac{\sin x\cos x}{\sin x+\cos x}dx$ on the top and bottom by $\cos x-\sin x$ to get

$\hspace{.6 in}\displaystyle\int\frac{\cos^2x\sin x}{2\cos^2 x-1}dx-\int\frac{\sin^2x\cos x}{1-2\sin^2 x}dx$.

Now substitute $u=\cos x$ in the first integral and $u=\sin x$ in the second integral to get

$\displaystyle\int\frac{u^2}{1-2u^2}du=\frac{1}{2}\int\big(-1+\frac{1}{1-2u^2}\big)du=\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]+C$
using partial fractions.

Now you can let $u=\cos x$ in the first term and $u=\sin x$ in the second to get

$\displaystyle\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}}-\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}$

---

Notice that substituting $t=\tan\frac{x}{2}$ gives

$\displaystyle\int{\frac{\sin x\cos x}{\sin x+ \cos x} dx=\int\frac{\frac{2t}{1+t^2}\cdot\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}}\cdot\frac{2}{1+t^2}dt=\int\frac{4t(1-t^2)}{(1+t^2)^2(1+2t-t^2)}dt$,

and now you can use partial fractions to continue.