Integration of TRIG (1 Viewer)

[skypryn]

Hi just a few questions one i got wrong in the half yearlys and can't find a fully worked proof in any textbook

primitive of cos(^2)2x

also, primitive of cos(^3)x

thanks heaps apart from that i'm alright at trig...

 

[:: ck ::]

mmm use ur cos2x = cos^2x - sin^2x
rearrange
shove cos^2x into the integrand

and u shud b able to go from there

with cos^3x... use substitution.. let u=sinx
so u end up with 1-u^2 in integrand
i.e sinx - sin^3x/3 + c

 

[Affinity]

[cos(2x)]^2 = (1/2)[1 + cos(4x)]
so it's primitive is (1/2)[x + (1/4)sin(4x)] + C

I will leave you to do the other: hint, expand cos(3x)

 

[kpq_sniper017]

just on the integration of cos³x.

is it faster to expand cos3x or to use a substitution.
i had the same question, and i did it both ways. IMO, the substitution was faster. any other opinions or objections?

 

[kpq_sniper017]

Originally posted by :: ryan.cck ::
mmm use ur cos2x = cos^2x - sin^2x
rearrange
shove cos^2x into the integrand

and u shud b able to go from there

with cos^3x... use substitution.. let u=sinx
so u end up with 1-u^2 in integrand
i.e sinx - sin^3x/3 + c

i don't think u would use that expansion of cos2x (correct me if i'm wrong). coz otherwise u'd just get cos2x - sin²x.
u'd probably have to use: cos2x = 2cos²x - 1.
but remember skypryn, ur finding the integral of cos²2x, so u'll have to change the double angle expansion slightly.

 

[:: ck ::]

Originally posted by pcx_demolition017
i don't think u would use that expansion of cos2x (correct me if i'm wrong). coz otherwise u'd just get cos2x - sin²x.
u'd probably have to use: cos2x = 2cos²x - 1.
but remember skypryn, ur finding the integral of cos²2x, so u'll have to change the double angle expansion slightly.

umm u make cos²x the subject... coz u can integrate functions in form trig(ax+b) as (1/a)(cotrig(ax+b) [obviousli being + or - depending on what ur integrating]

heres how i derive it........

Cos2x = 2cos²x-1
cos2x+1 = 2cos²x
cos²x = 1/2 (cos2x+1)

so integral of cos²x dx = 1/2integral(cos2x+1)dx
=1/2(1/2sin2x+x) + C
= (1/4)sin2x + (1/2) x + C

edit : CRAP didnt read the question.... cos^2 (2x)... lol ... read affinty's post

 

[skypryn]

yeah it's cool i get it and it's better to expand cos3x for other one as substitution without being asked is more like 4u.