Integral! Cal. 2!

  • Thread starter Whatupdoc
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  • #1
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[tex]\int 96cos^4(6x) * dx[/tex]

ok first i take out the 96 cause it's constant!

[tex] 96 \int cos^4(6x) * dx[/tex]

[tex] 96 \int (cos^2(6x)^2) * dx [/tex]

ok now with that setup, i can know use the half-angel formula!

[tex] 96 \int (\frac{1+cos(12x)}{2})^2 [/tex]

squared the problem...

[tex] 24 \int (1+ 2cos(12x) + (cos12x)^2 [/tex]

now to use the half-angle idents agian..

[tex] 24 \int 1 +2cos(12x) + 1/2(1+cos24x) [/tex]

can someone tell me if im doing this correctly before i integral the problem?
 
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Answers and Replies

  • #2
Pyrrhus
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Yes use Half Angle Identities.

[tex] cos^2(6x) = \frac{1+cos(12x)}{2} [/tex]

[tex]\int 96cos^4(6x) dx[/tex]

[tex] 96\int \frac{1+cos(12x)}{2} \frac{1+cos(12x)}{2}dx[/tex]

[tex] 96\int \frac{(1+cos(12x))^2}{4} dx[/tex]


[tex] 24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx][/tex]

You are doing fine...
 
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  • #3
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Cyclovenom said:
Yes use Half Angle Identities.

[tex] cos^2(6x) = \frac{1+cos(12x)}{2} [/tex]

[tex]\int 96cos^4(6x) dx[/tex]

[tex] 96\int \frac{1+cos(12x)}{2} \frac{1+cos(12x)}{2}dx[/tex]

[tex] 96\int \frac{(1+cos(12x))^2}{4} dx[/tex]

Expand and etc... keep going

Edit:Sleep got to me...

can i take out the 1/4? so 1/4*96 = 24
 
  • #4
Pyrrhus
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Whatupdoc said:
can i take out the 1/4? so 1/4*96 = 24
yes you can
 
  • #5
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thank you for checking my answer, ok i will cont. where i left off....


[tex] 24 \int 1 +2cos(12x) + 1/2(1+cos24x) [/tex]


[tex] 24 \int 1 +2cos(12x) + 1/2+\frac{1}{2}cos(24x) [/tex]


[tex]24 \int1dx + 48 \int cos(12x) + 12 \int dx + 12 \int cos(24x)[/tex]

adding alot of integral signs seems to look messy, but i think you can tell what im doing. ok time to find the anti-dervs.

[tex]24x + \frac{1}{12}(48)sin(12x) + 12 + 12(\frac{1}{24})sin(24x)[/tex]

k time to solve...
[tex]24x+4sin(12x) + 12 + \frac{1}{2}sin(24x)[/tex]

but it's the wrong answer
 
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  • #6
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Whatupdoc said:
expand this alittle more
[tex]1/2(1+cos24x) = 1/2+cos12x[/tex] right? <-- now sure about this part

This bit is wrong. 1/2(1 + cos 24x) = 1/2 + (1/2)cos 24x. (1/2)cos x isn't the same thing as cos (x/2).
 
  • #7
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Nylex said:
This bit is wrong. 1/2(1 + cos 24x) = 1/2 + (1/2)cos 24x. (1/2)cos x isn't the same thing as cos (x/2).

ah i see, dont know what i was thinking. i edited my previous post and fixed the problem, but the answer is still wrong
 
  • #8
Pyrrhus
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[tex] 24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx][/tex]

[tex] 24[\int 1 dx + \int 2cos(12x) dx + \int \frac{1+cos(24x)}{2} dx][/tex]

[tex] 24\int 1 dx + 24\int 2cos(12x) dx + 24\int \frac{1+cos(24x)}{2} dx [/tex]

[tex] 24\int 1 dx + 48\int cos(12x) dx + 12\int 1 dx+ 12\int cos(24x) dx [/tex]

[tex] 24x + 4sin(12x) + 12x+ \frac{sin(24x)}{2} + C [/tex]

[tex] 36x + 4sin(12x)+ \frac{sin(24x)}{2} + C[/tex]

Check my work...
 
  • #9
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Cyclovenom said:
[tex] 24[\int 1 dx + \int 2cos(12x) dx + \int \frac{1+cos(24x)}{2} dx][/tex]

[tex] 24\int 1 dx + 24\int 2cos(12x) dx + 24\int \frac{1+cos(24x)}{2} dx [/tex]
What in the world are you doing here!? Where did those "+ 24"s come from?

[edit]Nevermind, I was reading it wrong (since it looks cluttered).[/edit]
 
  • #10
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OK. I checked your last calculation. Everything is correct now.
 
  • #11
Pyrrhus
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Thanks but i meant that to whatuptdoc :smile:
 
  • #12
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ah i see what i did wrong, thank you
 

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