How will you integrate dx
1+sinxcosx
Remember that $\displaystyle \sin{2x} = 2\sin{x}\cos{x}$.
Therefore $\displaystyle \sin{x}\cos{x} = \frac{1}{2}\sin{2x}$.
So $\displaystyle \int{1 + \sin{x}\cos{x}\,dx} = \int{1 + \frac{1}{2}\sin{2x}\,dx}$
$\displaystyle = x - \frac{1}{4}\cos{2x} + C$.
Alternatively you could use a $\displaystyle u$ substitution.
$\displaystyle \int{1 + \sin{x}\cos{x}\,dx} = \int{1\,dx} + \int{\sin{x}\cos{x}\,dx}$
$\displaystyle = x + \int{\sin{x}\cos{x}\,dx}$.
Let $\displaystyle u = \sin{x}$ so that $\displaystyle \frac{du}{dx} = \cos{x}$.
Then you get
$\displaystyle = x + \int{u\,\frac{du}{dx}\,du}$
$\displaystyle = x + \int{u\,du}$
$\displaystyle = x + \frac{1}{2}u^2 + C$
$\displaystyle = x + \frac{1}{2}\sin^2{x} + C$
It's not difficult to show that the two answers are the same...
Learning something from the previous replies, you should be able to re-write the integral as $\displaystyle \int \frac{2}{2 + \sin (2x)} \, dx$.
Now read this: Wolfram|Alpha
Note that the standard Weierstrass substitution is used: PlanetMath: Weierstrass substitution formulas
You will find several examples if you search these forums using the search tool.
(There are other ways of doing this integral but they rely on clever algebraic parlor tricks).
$\displaystyle \int\frac{1}{1+\sin x\cos x}dx=\int\frac{1}{1+\frac{\sin 2x}{2}}dx=2\int\frac{dx}{2+\sin 2x}$
Now substitute $\displaystyle u=\tan x$.