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!
$\endgroup$ 12 Answers
$\begingroup$Hint:
With inside functions, in this case $(\sin 2x-\cos 2x)$, a $u$-substitution is a great place to start.
$\endgroup$ 2 $\begingroup$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 $x$.
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