Integrate $\int{\frac{\sin(2x)+\cos(2x)}{(\sin(2x)-\cos(2x))^{5/2}}}\,\mathrm dx$. 
Integrate $$\int{\frac{\sin(2x)+\cos(2x)}{(\sin(2x)-\cos(2x))^{5/2}}}\,\mathrm
 dx.$$

How do I integrate such an integral?
u-substitution, no idea
by parts, no idea
this is very confusing!  please help!
 A: Hint:
With inside functions, in this case $(\sin 2x-\cos 2x)$, a $u$-substitution is a great place to start.
A: One thing you can do when facing sums of trigonometric functions is to use the sum identities:
$$
\begin{eqnarray}
\sin (a + b) &= \sin a \cos b + \cos a \sin b \\
\cos (a + b) &= \cos a \cos b - \sin a \sin b
\end{eqnarray}
$$
In the expressions you have, sines and cosines have the same coefficient (essentially).  So, use $\sin \pi/4 = \cos \pi/4 = 1/\sqrt2$.
$$
\begin{eqnarray}
\sin 2x + \cos{2x}&= \sqrt2 \sin 2x \cos \pi/4 + \sqrt2 \cos2x \sin \pi/4\\
&=\sqrt2 \sin(2x + \pi/4)\\
\cos 2x - \sin{2x}&= \sqrt2 \cos 2x \cos \pi/4 - \sqrt2 \sin 2x \sin \pi/4\\
&= \sqrt2 \cos(2x + \pi/4)
\end{eqnarray}
$$
The integral becomes:
$$
\int{\frac{\sqrt2\sin(2x + \pi/4)}{(-\sqrt2\cos(2x + \pi/4))^{5/2}}}\
 dx
$$
Substitute $u = 2x + \pi/4$ and pull out the factors of $\sqrt2$ to get:
$$
\frac14 \int{\frac{\sin u}{(-\cos u)^{5/2}}}\
 du
$$
Let $v = -\cos u$:
$$
-\frac16 \int{\frac{dv}{v^{5/2}}}\
 du = -\frac16 v^{-\frac32} + C
$$
Now, substitute back for $u$, and then for $x$.
