cos^2 3x = 0.5(cos6x +1)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.
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.
COS6X = 2cos^3(x) - 1VenomP said:Sorry, this'll seem idiotic, but how did you work out the bolded line?
hilyounamu 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?lyounamu said:COS6X = 2cos^3(x) - 1
2cos^3(x) = cos6x + 1
cos^3(x)= 1/2(cos6x + 1)
hahahaha good one.L said:hi
im a tool
oh wait, no lol
thats you.
lol, didnt goxtcy said:hahahaha good one.
sup jamie, how was shopping?
you're manipulating the question to make it simple to integrate by using trig rules..[double angle cosine]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.