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!

  • 2
    $\begingroup$ You posted an erlier problem that was structurally almost the same as this one. It is important to use MSE answers to solve problems on your own. $\endgroup$ – André Nicolas Aug 9 '13 at 1:09


With inside functions, in this case $(\sin 2x-\cos 2x)$, a $u$-substitution is a great place to start.

  • $\begingroup$ Ok, so I should make u=sin2x-cos2x, then du=2cos2x+2sin2x=2(cos2x+sin2x) $\endgroup$ – user89632 Aug 9 '13 at 0:38
  • $\begingroup$ Oh that's right, thanks soo much! $\endgroup$ – user89632 Aug 9 '13 at 0:39

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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