How to solve that correctly?

[mellow condor]

I got pi/3

and must get 5/8 * pi

[mellow condor]

what i did is

1 + cos (6*theta)
----------------- ( divided )
6

and then there are two integrals bla bla bla

2pi + sen (6 theta)
------------- ( must do sarrus from 0 to 2pi that makes that 0
6

then 2pi / 6 sooo = pi/3

[prisma osprey]

Hi

[prisma osprey]

If you used the double angle formula

cos^2(theta) = (1 + cos2theta)/2

You con't just use 6 instead of 2

What you should do is rewrite:
cos^6(theta) as (cos^2(theta))^3 then use the formula giving you

((1 + cos2theta)/2)^3

[mellow condor]

no way

[mellow condor]

i went from this to cos 6 because teacher told

how could i solve that

[prisma osprey]

Nah
I'm pretty sure you can't put cos(6theta) unless he had another thing in mind.

To solve this first expand (1 + cos(2theta))^3

[mellow condor]

how

[prisma osprey]

You could just do
(1 + cos(2theta)) * (1 + cos(2theta)) * (1 + cos(2theta))

Or use the identity
(a + b)^3 = a^3 + 3a^2.b + 3a.b^2 + b^3

So
(1 + cos2theta)^3 = 1 + 3cos(2theta) + 3cos^2(2theta) + cos^3(2theta)

[mellow condor]

i did all the math

and got to here

Sorry for being late

I had class today and asked teacher, didn't see your answer

teached told me to do that too

WHAT

I can underestand why

down part is equal to the upper part

but, because its doing upper part and the down one is negative, area =0

but if we do individually

and sum them

won't be 0

because area cannot be negative

[prisma osprey]

Hi
So
Don't worry for the late response.

1. Yes when talking about integral we actually are talking about signed Areas. So Yes the area can in fact be negative which is the reason why the integral from 0 to 2pi of cos2x = 0

[mellow condor]

oh

well

at least my answer is correct?