Integral [0 to pi] [sin(2x)*cos(3x)*sin(4x)*cos(5x)*sin(6x)] dx

[Vali]

Hi!

. . . . .\(\displaystyle \displaystyle \int_0^{\pi}\, \left[\, \sin(2x)\, \cdot\, \cos(3x)\, \cdot\, \sin(4x)\, \cdot\, \cos(5x)\, \cdot\, \sin(6x)\, \right]\, dx\)

I tried to group the terms, then to transform in sum but didn't work.
Some ideas?

 

[sinx]

Hi!

. . . . .\(\displaystyle \displaystyle \int_0^{\pi}\, \left[\, \sin(2x)\, \cdot\, \cos(3x)\, \cdot\, \sin(4x)\, \cdot\, \cos(5x)\, \cdot\, \sin(6x)\, \right]\, dx\)

I tried to group the terms, then to transform in sum but didn't work.
Some ideas?

have you tried trig identity, double angle?
sub u=2x?
can you show what you have tried?

 

[Dr.Peterson]

Hi!

. . . . .\(\displaystyle \displaystyle \int_0^{\pi}\, \left[\, \sin(2x)\, \cdot\, \cos(3x)\, \cdot\, \sin(4x)\, \cdot\, \cos(5x)\, \cdot\, \sin(6x)\, \right]\, dx\)

I tried to group the terms, then to transform in sum but didn't work.
Some ideas?

Since this is a definite integral, my first thought was to wonder if there is something special about the interval. In particular, maybe over [0,pi] there might be a symmetry that would result in positive and negative areas canceling out, yielding a result of 0. Or there might be some other similar effect.

To determine that, I might call the integrand f(x), and find out what f(pi - x) is.

 

[Steven G]

Since this is a definite integral, my first thought was to wonder if there is something special about the interval. In particular, maybe over [0,pi] there might be a symmetry that would result in positive and negative areas canceling out, yielding a result of 0. Or there might be some other similar effect.

To determine that, I might call the integrand f(x), and find out what f(pi - x) is.

Funny as i was just thinking the same thing!

 

[Vali]

Thank you for suggestions.I calculated f(pi-x) and it gave -f(x)

 

[Dr.Peterson]

Thank you for suggestions.I calculated f(pi-x) and it gave -f(x)

Good. Now do you see what that implies about the integral?

 

[Deleted member 4993]

Hi!

. . . . .\(\displaystyle \displaystyle \int_0^{\pi}\, \left[\, \sin(2x)\, \cdot\, \cos(3x)\, \cdot\, \sin(4x)\, \cdot\, \cos(5x)\, \cdot\, \sin(6x)\, \right]\, dx\)

I tried to group the terms, then to transform in sum but didn't work.
Some ideas?

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[/FONT]f(pi/2-x) = sin(pi - 2x) * cos(3pi/2 - 3x) * sin(2pi - 4x) * cos(5pi/2 - 5x) * sin(3pi - 6x)

=[sin(2x)] * [-sin(3x)] * [-sin(4x)] * [sin(5x)] * [sin(6x)] = {sin(2x) * sin(3x) * sin(4x) * sin(5x) * sin(6x)}

f(pi/2+x) = sin(pi + 2x) * cos(3pi/2 + 3x) * sin(2pi + 4x) * cos(5pi/2 + 5x) * sin(3pi + 6x)

=[-sin(2x)] * [sin(3x)] * [sin(4x)] * [-sin(5x)] * [-sin(6x)] = -{sin(2x) * sin(3x) * sin(4x) * sin(5x) * sin(6x)}

So:

f(pi/2-x) = - f(pi/2+x)

So what does it tell about the \(\displaystyle \int_0^{\pi}\)

 

[Vali]

Is it right ?

 

[Dr.Peterson]

Is it right ?

There are several ways you could have expressed it; this is a very nice one. Good work.