Evaluation of $\int \sqrt{1+\cot x}dx$? 
What is $$\int \sqrt{1+\cot x}dx$$  

My friends and I tried using all possible trigonometric formula. We couldn't find a way to solve it 
Please help me solve it.

Put $(1+cotx)=t^2$ in the given integral and then do the reqd. Substitutions to get :
$$\int  -\frac {2x^2dx}{x^4-2x^2+2}$$
I think we can avoid complex numbers.
Write it as:
$$\int -2(\frac {(x^2+ \sqrt2+x^2-\sqrt2)dx}{x^4-2x^2+2})$$
Split it into two integrals,
$$\int\frac {(x^2+\sqrt2)dx}{x^4-2x^2+2}$$
And
$$\int \frac {(x^2-\sqrt2)dx}{x^4-2x^2+2}$$
Simply divide by $x^2$ in both the integrals to get,
$$\int \frac {(1+\frac {\sqrt2}{x^2})dx}{x^2+2/x^2-2}$$
Now complete the square in the denominator as $(x-\sqrt2/x)^2 + 2(\sqrt2-1)$ .    (for first integral)
Then substitute $x-\sqrt(2)/x = v$.
To get the integrals as
$$\int \frac{dv}{(v^2+ 2(\sqrt(2)-1))}$$
And $$\int \frac{dv}{(v^2-2(\sqrt(2)+1))}$$
Which are standard integrals.
