Integration with half angle identity

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mikewill54

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Hi,
Im having trouble with this question [math]\int cos^2xcos(5x) dx[/math]Can i simply split them into two separate integrals? and integrate them both separately, using a half angle identity for the first part
Thanks for any help
Mike
 
mikewill54 said:
Hi,
Im having trouble with this question [math]\int cos^2xcos(5x) dx[/math]Can i simply split them into two separate integrals? and integrate them both separately, using a half angle identity for the first part
Thanks for any help
Mike
Click to expand...
Please show us how do you propose to split the integral. There are several ways you can achieve that.
 
mikewill54 said:
Hi,
Im having trouble with this question [math]\int cos^2xcos(5x) dx[/math]Can i simply split them into two separate integrals? and integrate them both separately, using a half angle identity for the first part
Thanks for any help
Mike
Click to expand...
I think what you're suggesting is
[math]\int \cos^2(x) \cos(5x)\, dx\neq\int\cos^2(x) \, dx\cdot \int \cos(5x)\, dx[/math]If so, this would be incorrect.
I would rewrite the integrand using the following identities, and then split the integrals:
[math] 1)\quad \cos^2(\alpha)=\frac{1}{2}(\cos(2\alpha)+1)\\ 2)\quad \cos(\alpha)\cos(\beta)=\frac{1}{2}\left[\cos(\beta+\alpha)+\cos(\beta-\alpha)\right] [/math]
 
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Hi
Thanks for all the help, is it possible to do it like this,


[math]\int cos^2(x)cos(5x)dx \\= \int \frac{1+cos(2x)}{2}dx+\int cos(5x)dx \\= \frac{1}{2} \int 1+cos(2x)dx+ \int cos (5x)dx \\= \frac{1}{2}(\int(1dx+ \int(cos(2x)dx)+\int cos(5x)dx \\= \underline {\frac{1}{2}(x)} \\ let \ u=2x \ \\ \frac{du}{dx}=2, \ \ dx= \frac{1}{2}du \\ = \int cos(u)du \ = sin(u) \\= \underline {\frac{1}{2}sin(2x)} \\=let \ u = 5x \\= \frac{du}{dx}=5, \ dx=\frac{1}{5}du \\ = \int cos(u)du \ = sin(u) \\= \underline {\frac{1}{5}sin(5x)}[/math]
 
mikewill54 said:
Hi
Thanks for all the help, is it possible to do it like this,


[math]\int cos^2(x)cos(5x)dx \\= \int \frac{1+cos(2x)}{2}dx+\int cos(5x)dx \\= \frac{1}{2} \int 1+cos(2x)dx+ \int cos (5x)dx[/math]
Click to expand...
Since when is [imath]\int A(x) B(x) ~ dx = \int A(x) ~ dx + \int B(x) ~ dx[/imath]?

[math]\int cos^2(x) ~ cos(5x) ~ dx = \int \dfrac{1}{2} (1 + cos(2x) ) ~ cos(5x) ~ dx = \dfrac{1}{2} \int (cos(5x) + cos(2x) ~ cos(5x) ) ~ dx[/math]
The first integration is trivial. Do you know a trig identity that will help do the second integration?

-Dan

Addendum:
Do you have to use the half angle substitution? Frankly I find it more straightforward to consider that [imath]Re(e^{5ix}) = cos(5x)[/imath] so [math]\int cos^2(x) ~ cos(5x) ~ dx = Re \left ( \int cos^2(x) e^{5ix} ~ dx \right )[/math]and do integration by parts.
 
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topsquark said:
Since when is [imath]\int A(x) B(x) ~ dx = \int A(x) ~ dx + \int B(x) ~ dx[/imath]?

[math]\int cos^2(x) ~ cos(5x) ~ dx = \int \dfrac{1}{2} (1 + cos(2x) ) ~ cos(5x) ~ dx = \dfrac{1}{2} \int (cos(5x) + cos(2x) ~ cos(5x) ) ~ dx[/math]
The first integration is trivial. Do you know a trig identity that will help do the second integration?

-Dan
Click to expand...
I gave the identity for the second integration in post#3 :)
 
mikewill54 said:
Hi
Thanks for all the help, is it possible to do it like this,


[math]\int cos^2(x)cos(5x)dx \\= \int \frac{1+cos(2x)}{2}dx+\int cos(5x)dx \\= \frac{1}{2} \int 1+cos(2x)dx+ \int cos (5x)dx \\= \frac{1}{2}(\int(1dx+ \int(cos(2x)dx)+\int cos(5x)dx \\= \underline {\frac{1}{2}(x)} \\ let \ u=2x \ \\ \frac{du}{dx}=2, \ \ dx= \frac{1}{2}du \\ = \int cos(u)du \ = sin(u) \\= \underline {\frac{1}{2}sin(2x)} \\=let \ u = 5x \\= \frac{du}{dx}=5, \ dx=\frac{1}{5}du \\ = \int cos(u)du \ = sin(u) \\= \underline {\frac{1}{5}sin(5x)}[/math]
Click to expand...
NO!

In general

\(\displaystyle \int f(x)* g(x) dx \)

DOES NOT equal to

\(\displaystyle \int f(x) dx + \int g(x) dx \)a

Check the second line of your response !!
 
Ok, really appreciate all the help... I think I've got it this time

[math]cos^2xcos(5x)dx \\Rule 1 \\ 1. Cos^2(x)= (\frac{1}{2}cos(2x)+1) \\ \int (\frac{1}{2}(1+cos(2x)cos(5x))dx \\ \frac{1}{2} \int ((1+cos(2x))cos(5x)dx \\ Multiply \ out \ the \ brackets \\ \frac{1}{2} \int (cos(5x)+cos(2x)cos(5x)dx \\ Rule 2 \\ Cos(x)Cos(y)= \frac{1}{2}[cos(x+y)+cos(x-y)] \\ \frac{1}{2} \int cos(5x)+ \frac{1}{2}[cos(5x+2x)+cos(5x-2x)] \\ \frac{1}{2} \int cos(5x)+ \frac{1}{2}[cos(7x)+cos(3x)] \\ \frac{1}{2} \int cos(5x)+ \int \frac{1}{2}cos(7x)+ \int \frac{1}{2}cos(3x) \\ \underline{\frac{1}{10}sin(5x)+\frac{1}{14}sin(7x)+\frac{1}{6}sin(3x)}[/math]
 
[math]\red{\frac{1}{2}} \int cos(5x)+ \frac{1}{2}[cos(7x)+cos(3x)][/math]There's one mistake here. The 1/2 in the front needs to get distribute to all the terms, not just the first one. Also, don't forget your constant of integration.
 
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