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Integration q involving squared sign (1 Viewer)

VenomP

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int. from pi/4 to 0 of cos^2 3x

What do you do here? My solutions look totally odd.

Is it a product rule? Would it be easier to use a chain rule or something?

Thanks.
 

shaon0

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VenomP said:
int. from pi/4 to 0 of cos^2 3x

What do you do here? My solutions look totally odd.

Is it a product rule? Would it be easier to use a chain rule or something?

Thanks.
cos^2 3x = 0.5(cos6x +1)
S cos^2 3x
= 0.5 S cos6x dx + 0.5 S 1 dx
= 1/12 sin6x + x/2 +C
Sub in the limits now.

Sorry its not that thorough. I'm a little busy.
 
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VenomP

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shaon0 said:
cos^2 3x = 0.5(cos6x +1)
S cos^2 3x
= 0.5 S cos6x dx + 0.5 S 1 dx
= 1/12 sin6x + x/2 +C
Sub in the limits now.

Sorry its not that thorough. I'm a little busy.

Sorry, this'll seem idiotic, but how did you work out the bolded line?
 

lyounamu

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VenomP said:
Sorry, this'll seem idiotic, but how did you work out the bolded line?
COS6X = 2cos^3(x) - 1
2cos^3(x) = cos6x + 1
cos^3(x)= 1/2(cos6x + 1)
 

VenomP

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lyounamu said:
COS6X = 2cos^3(x) - 1
2cos^3(x) = cos6x + 1
cos^3(x)= 1/2(cos6x + 1)
What does 2cos^3(x) have to do with anything when the question involves cos^2 3x?

I'm not being mean here, even though it sounds it. I really do appreciate your help.
 

SeftonIsAHole

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VenomP said:
What does 2cos^3(x) have to do with anything when the question involves cos^2 3x?

I'm not being mean here, even though it sounds it. I really do appreciate your help.
you're manipulating the question to make it simple to integrate by using trig rules..[double angle cosine]
 

addikaye03

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Since cos2x=2cos^2x-1 OR 1-2sin^2x, therefore cos4x=2cos^2(2x)-1, then cos8x=2cos^2(4x)-1 therefore cos6x=2cos^2(3x)-1, by rearranging 2cos^2(3x)=cos6x+1 and cos^2(3x)=0.5(cos6x+1), then int. accordingly

Remember the process, dont remember the transformation, theres already enough shit to remember
 

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