Calculus - Anti-Derivatives
posted by Sean .
How would you find the
Integral of (cos(x\8))^3, defined from
-(4 x pi)/3) to (4 x pi)/3)
cos^3(x)dx = cos^2(x)cos(x)dx =
Is cos(x)8 supposed to be
cos 8x, (cosx)^8 or 8 cosx?
Its actually supposed to be (cos(x/8))^3.
With what Anonymous answered, I see he left out the x/8, instead only using x. Would you, Drwls, mind using x/8?
Sorry, I missed the \ mark. It is always better to use / for fractions when typing.
Without messing with trig identities for cos (x/8), let's just substitute u for x/8. Then your integral becomes
Integral of (cos u)^3
= (1- sin^2 u) cos u du , from
u = -(pi/6) to (pi/6)
Now let sin u = v
dv = cos u du
and your integral becomes
Integral of (1- v^2)dv ,
from v = -1/2 to 1/2,
since that is what v is when u = +/- pi/6
Now it should be easy!
The indefinite integral is v - v^3/3. Evaluate it at v = 1/2 and -1/2.
My answer: 11/24 - (-11/24) = 11/12
Don't trust my algebra. Check it yourself.
I can follow everything except when you change the (4 x pi)/3 to pi/6, and change pi/6 into 1/2. Could you clear that up for me?
the upper limit for example was
x = 4 pi/3
u = x/8
so the upper limit using u instead of x is
u = (4 pi/3) /8
u = pi/6
then it changes again to go from u to v
v = sin u
so at that same upper limit where x = 4 pi/3 and u = pi/6, we need to find v
v at this upper limit is:
v = sin (pi/6) = sin 30 degrees = 1/2
Oh. Thank you, Damon, and Drwls. That really helped me out.