Calculate integrals $\int_{{\pi \over 4}}^{\arctan {1 \over 2}} {{{\sqrt {{\mathop{\rm cos}\nolimits u} } } \over {\sqrt {\sin u} }}du} $ This is my homework. And it's really a challenge for me. Can anyone solve this.

$$\large\int\limits_{{\pi  \over 4}}^{\arctan {1 \over 2}} {{\sqrt{\cos u}   \over {\sqrt {\sin u} }}du} $$

 A: You can simplify the integral as follows:
$$
\int \sqrt{\cot u}~ du
$$
Let's set $$I_1=\int (\sqrt{\tan u}+\sqrt{\cot u})~du$$
By taking $u=\arctan t^2$ so $du= \frac{2t}{1+t^4}dt$ and then:
 $$I_1=2\int \frac{t(t+\frac{1}{t})}{1+t^4}dt=2\int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt$$
Now, setting $t-\frac{1}{t}=z$, we get $(1+\frac{1}{t^2})dt=dz$ and $I_1$ would be changed to:
$$I_1=2\int \frac{1}{z^2+2}dz =\sqrt{2}\tan^{-1}\left(\frac{z}{\sqrt{2}}\right)+c=\sqrt{2}\tan^{-1}\left(\frac{\sqrt{\tan u}-\sqrt{\cot u}}{\sqrt{2}}\right)+C$$
By the similar way, if we take $$I_2=\int (\sqrt{\tan u}-\sqrt{\cot u})du$$
then, in denominator make term $(t+1/t)^2-1$ and substitute $t+1/t=z$ next. This gives us $I_2$. Now we have:
$$I_3=\int \sqrt{\cot u}du=\frac{I_1-I_2}{2}$$
I let you put the bounds and calculate the value!
A: \begin{equation}
\begin{split}
I=&\int\sqrt{\cot u} \ du \\
\ =&\sqrt{2}\int \frac{\cos u\ du}{\sqrt{\sin 2u}}\\
\ =&I_1+I_2
\end{split}
\end{equation} where $$I_1=\frac{1}{\sqrt{2}}\int\frac{\cos u-\sin u}{\sqrt{(\cos u+\sin u)^2-1}}du$$ and $$I_2=\frac{1}{\sqrt{2}}\int\frac{\cos u+\sin u}{\sqrt{1-(\sin u-\cos u)^2}}du$$
Since $$\frac{d}{du}(\cos u+\sin u)=\sin u-\cos u$$ and $$\frac{d}{du}(\sin u-\cos u)=\cos u+\sin u$$ we get after replacing $\cos u+\sin u$ by $z$ and $-\cos u+\sin u$ by $t$ $$I_1=\frac{1}{\sqrt{2}}\int \frac{dz}{\sqrt{z^2-1}}=\frac{1}{\sqrt{2}}\ln(z+\sqrt{z^2-1})+c_1\\=\frac{1}{\sqrt{2}}\ln(\cos u+\sin u+\sqrt{\sin 2u})+c_1$$ and $$I_2=\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{1-t^2}}=\frac{1}{\sqrt{2}}\arcsin(t)+c_2=\frac{1}{\sqrt{2}}\arcsin(\sin u-\cos u)+c_2$$
Now, you can evaluate the integral by putting the bounds. 
