What is the integration of $\sqrt {\tan x+\cot x}$? I saw a question about what would be the integration of $\sqrt {\tan x}+\sqrt{\cot x}$!  Well, it was not that tough.
But after that I thought what would be the integral of $\sqrt {\tan x+\cot x}$. And with my mathematical knowledge I couldn't solve it. (I am a 12th grader)
When I tried to do it the wolfram integral calculator, it did give some result but which was out of my grasp(Hyper geometric function!!!). Here is the link:
http://integrals.wolfram.com/index.jsp?expr=Sqrt%5Bcot+x%2Btanx%5D&random=false
So can you give me an answer that I can grasp?
 A: Many integrals can not be solved using elementary functions such as powers, trigonometry and exponentials and their inverses. For some such integrals which come up often Mathematicians have made up new functions which are defined in terms of the integral. These functions can still be studied and their properties examined and other applications found.
The Hypergeometric function is an example of such a function. It can be defined in several different ways (including infinite sums as well as integrals). A definition involving integrals is:
$$_2F_1(a,b;c;x)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(b-c)}\int_0^1\frac{t^{b-1}(1-t)^{c-b-1}}{(1-tx)^a}dt$$
, where $\Gamma(x)$ is the Gamma function which is like factorial but works on all numbers not just natural numbers.
So your result of:
$$-\frac{2 _2F_1\left(\frac14,\frac34;\frac54;\cos^2x\right)}{\sqrt[4]{\sin^2x}\sqrt{\csc x\sec x}}$$
could be written as:
$$-\frac{2 \frac{\Gamma(\frac54)}{\Gamma(\frac34)\Gamma(-\frac12)}\int_0^1\frac{1}{t^{1/4}(1-t)^{1/2}(1-\cos^2x\cdot t)^{1/4}}dt}{\sqrt[4]{\sin^2x}\sqrt{\csc x\sec x}}$$
A: I can give you a rather simple answer if you consider a definite integral instead of an indefinite integral (a primitive) that looks very complicated. 
In fact a small calculation gives $\sqrt{\tan x+\cot x}=1/\sqrt{\cos(x) \sin(x)}$. 
Taking bounds that are natural (tangent is not defined in $\pi/2$, cotangent is not defined in $0$)
$$\tag{1}I=\int_0^{\pi/2}\cos(x)^{-1/2} \sin(x)^{-1/2}dx$$
This is the area of the region under the curve represented below. Though not bounded, its value is finite as we are going to see it.

Now, have a look at (https://en.wikipedia.org/wiki/Beta_function) where the $Beta$ integral is defined. It is a rather important tool, extending to a continuous context the - inverse of - binomial coefficients, whence their names). The general Beta integral depends on two parameters ; it may be defined by: 
$\tag{2}B(x,y)=2\int_0^{\pi/2}\cos(x)^{2x-1} \sin(x)^{2y-1}dx$
The connection between (1) and (2) is made by taking $x=y=1/4$,  giving: 
$$I=\dfrac{1}{2}B\left(\dfrac{1}{4},\dfrac{1}{4}\right)\approx3.70815$$
(the last value has been obtained with Mathematica ; in fact, all scientific computer software are able to compute this kind of integral).
Remark : one may wonder how this $B\left(\dfrac{1}{4},\dfrac{1}{4}\right)$ can be computed ; it is "very simple":
$$B\left(\dfrac{1}{4},\dfrac{1}{4}\right)=\dfrac{(-3/4)! \times (-3/4)!}{(-1/2)!}$$
where the "!" designates an extension of the classical "factorial" symbol.
A last remark: $(-1/2)!=\sqrt{\pi}$. Astonishing, no ? In fact, it is the Gamma function which is hidden behind the factorial notation. I advise you to learn what this function is, and what are its properties (https://en.wikipedia.org/wiki/Gamma_function): it is not difficult, and may be very rewarding.
Wishing, as you are a 12th grader, that it has not been too difficult, and maybe that it has given you some appetite for your forthcoming maths !
A: tan x + cot x = t^2. (tan x)^2 + (cot x)^2 + 2 = t^4.
((sec x )^2 + (cosec x)^2)dx = 2tdt = (2 + (tan x)^2 + (cot x)^2 )dx = t^4 dx.
You can substitute and simplify from here.
