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# Quick Integration (P4) watch

1. Hi,

just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

Integrate: 4 . cos^2[3x]

Many Thanks

Streety
2. cos6x = 2cos^2(3x) - 1
3. (Original post by streetyfatb)
Hi,

just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

Integrate: 4 . cos^2[3x]

Many Thanks

Streety
cos2x = cos^2x - sin^2x = cos^2x - (1-cos^2x) = 2cos^2x - 1
cos^2(x) = 0.5(cos2x + 1)
therefore
cos^2(3x) = 0.5(cos6x + 1)
4cos^2[3x] =2cos6x + 2
4. (Original post by streetyfatb)
Hi,

just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

Integrate: 4 . cos^2[3x]

Many Thanks

Streety
Hello,

I just joined and came across your question; i havn't done it yet, but i think the idea is to use the cos idedtity:

cos 2x = 2cos^2 x - 1

letting x = 3x, which gives teh LHS as (cos 6x +1)/2 as being = 2 cos^2 x

i can't see anything wrong with that, lol
5. (Original post by streetyfatb)
Hi,

just unsure on how to deal with this integral as it deals with 3x instead of 2x, which is what I am used to, and do not know the identities needed.

Integrate: 4 . cos^2[3x]

Many Thanks

Streety
Hello,

I just joined and came across your question; i havn't done it yet, but i think the idea is to use the cos idedtity:

cos 2x = 2cos^2 x - 1

letting x = 3x, which gives teh LHS as (cos 6x +1)/2 as being = 2 cos^2 3x

i think that the idea
6. déja vu

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