Getting nowhere with this trigonometric integration: $\left(\frac{1}{1 + \sin x \cos{x}}\right)$ guys I need help integrating the function. I have been thinking about it for an hour and got nowhere.
$$\left(\frac{1}{1 + \sin x \cos{x}}\right)$$
 A: HINT:
$$1+\sin x\cos x=\cos^2x(1+\tan^2x+\tan x)=\dfrac{1+\tan^2x+\tan x}{\sec^2x}$$
Set $\tan x=y$
More generally, for $$A\sin^2x+B\cos^2x+C\sin x\cos x+D$$ take $\cos^2x$ or $\sin^2x$
A: Hint 
Substitute $\tan \frac{x}{2}= t$.
A: Well, we have, $$I = \int \frac {1}{1+ \sin x \cos x} \mathrm {d}x $$ $$= \int \frac {2}{\sin 2x +2} \mathrm{d}x $$ $$= \int \frac {2}{\frac {2\tan x}{1+\tan^2 x} + 2} \mathrm {d}x $$ Substituting $u = \tan x $, we get, $$I = \int \frac {1}{u^2+ u+1} \mathrm {d}u $$ $$= \int \frac{1}{( u + \frac {1}{2})^2 + (\frac {\sqrt {3}}{2})^2} \mathrm {d}u $$ Hope you can take it from here. 
A: An alternative approach, which always works with rational functions containing polynomials of $\sin(x)$ and $\cos(x)$.
First, use Euler's identity $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$  and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$.Now, substitute $u=e^{ix}$, $u^{-1}=e^{-ix}$ and $du=ie^{ix}dx \implies dx =\frac{du}{iu}$. The resulting integral is always a rational function containing only polynomials in $u$. The resulting integral can then be solved by partial fractions.
