2
$\begingroup$

I need some help solving this integral, seems nasty to me!

$$ \int e^{-x}\cos4x\cos2x\,\mathrm dx $$

I tried integration by parts, but that seemed to me of no use.

There's also a similar one, maybe it might help solving this one or vice versa.

$$ \int x\sin x\sin 2x\sin 3x\,\mathrm dx $$

Again, please try to give just hints...! (As I always ask :D)

$\endgroup$
  • $\begingroup$ Are you familiar with Euler's formula? $\endgroup$ – copper.hat Apr 19 '13 at 14:20
  • $\begingroup$ Yes, but so far we've not used complex numbers in solving integrals. Although bring it on - let's see how useful complex numbers can be in solving such integrals! $\endgroup$ – Parth Thakkar Apr 19 '13 at 14:21
3
$\begingroup$

Use the fact that

$$2 \cos{a x} \cos{b x} = \cos{(a-b) x} + \cos{(a+b) x}$$

and

$$\int dx\: e^{p x} \cos{q x} = \frac{p \cos{q x} + q \sin{q x}}{p^2+q^2} e^{p x} + C$$

where $C$ is a constant of integration.

$\endgroup$
  • $\begingroup$ Quite funny! I almost forgot about the 2nd formula! $\endgroup$ – Parth Thakkar Apr 19 '13 at 14:35
1
$\begingroup$

$\cos4x\cos2x=\frac{1}{2}(\cos6x+\cos2x)$


Another method is to use $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

$\endgroup$
1
$\begingroup$

HINT: $2\cos{x}*\cos{y}=\cos{(x-y)}+\cos{(x+y)}$
Next, use integration by parts twice.

$\endgroup$
  • $\begingroup$ Ok, that seems to be working! Although the answer turns out to be humongous. Never mind, I got the answer at least! $\endgroup$ – Parth Thakkar Apr 19 '13 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.