lp_06
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#1
Report Thread starter 13 years ago
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i know its easy but im stuck on how to integrate cos2xcosx. do i have to use cos2x=1-2sin^2(x)
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investor
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u wil have to use intergration by parts and dv/dx as cos2x and then v=(sin2x)/2 and u =cos du/dx = -sin so it will be something like

(cosxsin2x)/2 - integral of -(sinxsin2x)/2
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Knogle
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\cos2xcosx = (2\cos^2x - 1)cosx = 2\cos^3x - cosx

You can integrate that now, can't you?
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Bengaltiger
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Actually, its not that easy.

cos3x = cos (2x + x) = cos2xcosx - sin2xsinx

cosx = cos ( 2x - x) = cos2xcosx + sinx2xsinx

Adding both gives

cos3x + cosx = 2cos2xcosx

Therefore You can intergrate 0.5 (cos3x + cosx)

which is (1/6)sin3x + (1/2)sinx + c
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lp_06
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cheers i get it now, thanks
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username9816
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(Original post by lp_06)
i know its easy but im stuck on how to integrate cos2xcosx. do i have to use cos2x=1-2sin^2(x)
cos2x.cosx = [1 - 2sin^2(x)].cosx = cosx - 2cosx.sin^2(x) = cosx - 2sin^2(x).[d/dx(sinx)]
Recall: Integral f ' (x).[f(x)]^n dx = [f(x)]^(n + 1) / (n + 1) + k
:. Integral cos2x.cosx dx = Integral cosx - 2sin^2(x).[d/dx(sinx)] dx = sinx - 2sin^3(x)/3 + k = sinx[1 - (2/3)sin^2(x)] + k
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