integration (1 Viewer)

[want2beSMART]

the integral of cos5xcos4xdx

i made cos4x into (1-sin^2x)^2

dont know what to do next

 

[KFunk]

2cos5xcos4x = cosx + cos9x

∫cos5xcos4xdx
= 1/2 ∫ cosx + cos9x dx
= 1/2sinx +1/18.sin9x +c

 

[want2beSMART]

how did you get 1/2 ∫ cosx + cos9x dx

 

[Slidey]

cos5xcos4xdx
cosa+cosb=cos((a+b)/2)cos((a-b)/2)
a=9
b=1

 

[KFunk]

By using the sums as products thing. Consider the following:

(1) cos(θ + φ ) = cosθcosφ - sinθsinφ
(2) cos(θ - φ ) = cosθcosφ + sinθsinφ

by adding (1) and (2)

cos (θ + φ ) + cos (θ - φ ) = 2cosθcosφ

i.e 2cosθcosφ = cos(θ + φ ) + cos(θ - φ ) which translates directly to the above.

2cos5xcos4x = cos(5x + 4x) + cos(5x - 4x)
cos5xcos4x = 1/2(cosx + cos9x)

hence ∫ cos5xcos4x dx = 1/2 ∫ cosx + cos9x dx

 

[want2beSMART]

i understand very well now

thank you very much