Evaluation of $\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$ Evaluate the given integral:

$$\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$$

After using $\int_{a}^{b}f(x)dx= \int_{a}^{b} f(a+b-x)dx$, we get $\int_{\pi/6}^{\pi/3} \sqrt{\sin(x)+\cos(x)} dx$ but I am not able to solve this integral. Could someone help me with this?
 A: Hint:  Let $u=\sqrt{1+\tan x}$, so $du=\frac{\sec^2x}{2\sqrt{1+\tan x}}dx\;$.  Then
$\displaystyle\int\frac{1}{\sqrt{1+\tan x}}dx=2\int\frac{1}{1+\tan^2x}\frac{\sec^2 x}{2\sqrt{1+\tan x}}dx=2\int\frac{1}{1+(u^2-1)^2}du=2\int\frac{1}{u^4-2u^2+2}du$.  
Next use partial fractions:

Since $u^4-2u^2+2=[(u^2)^2+2\sqrt{2}u^2+(\sqrt{2})^2]-(2+2\sqrt{2})u^2=(u^2+\sqrt{2})^2-\left(\sqrt{2+2\sqrt{2}}u\right)^2$
$\hspace{1.32 in}=\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$,
$\hspace{.4 in}\displaystyle\frac{1}{u^4-2u^2+2}=\frac{Au+B}{u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}}+\frac{Cu+D}{u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}}$ so
$\hspace{.3 in}1=(Au+B)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)+(Cu+D)\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$.
1) $u=0\;$ gives $B+D=\frac{\sqrt{2}}{2}$
2) the $u^3$ coefficient gives $0=A+C$
3) the $u$ coefficient gives $0=\sqrt{2}(A+C)+\sqrt{2+2\sqrt{2}}(B-D)$, so $\color{blue}{B=D=\frac{\sqrt{2}}{4}}$
4) the $u^2$ coefficient gives $0=\sqrt{2+2\sqrt{2}}(A-C)+(B+D)=\sqrt{2+2\sqrt{2}}(A-C)+\frac{\sqrt{2}}{2}$,
so $A-C=-\frac{1}{2\sqrt{1+\sqrt{2}}}=-\frac{\sqrt{\sqrt{2}-1}}{2}$ and $\color{blue}{A=-\frac{\sqrt{\sqrt{2}-1}}{4}}$ and $\color{blue}{C=\frac{\sqrt{\sqrt{2}-1}}{4}}$.
(Now integrate each term.)
A: hint: $\sin x + \cos x = \sqrt{2}\sin(x+\frac{\pi}{4})$, and you get an Elliptic Integral of Second Kind. This part has been posted here at MSE.
