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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)

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Are you familiar with Euler's formula? –  copper.hat Apr 19 '13 at 14:20
    
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! –  Parth Thakkar Apr 19 '13 at 14:21
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3 Answers

up vote 3 down vote accepted

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.

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Quite funny! I almost forgot about the 2nd formula! –  Parth Thakkar Apr 19 '13 at 14:35
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$\cos4x\cos2x=\frac{1}{2}(\cos6x+\cos2x)$


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

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

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