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Calculus - Anti-Derivatives

posted by on .

How would you find the

Integral of (cos(x\8))^3, defined from

-(4 x pi)/3) to (4 x pi)/3)

  • Calculus - Anti-Derivatives - ,

    cos^3(x)dx = cos^2(x)cos(x)dx =

    [1-sin^2(x)]cos(x)dx =

    [1-sin^2(x)]d(sin(x))

  • Calculus - Anti-Derivatives - ,

    Is cos(x)8 supposed to be
    cos 8x, (cosx)^8 or 8 cosx?

  • Calculus - Anti-Derivatives - ,

    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?

  • Calculus - Anti-Derivatives - ,

    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.

  • Calculus - Anti-Derivatives - ,

    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?

  • Calculus - Anti-Derivatives - ,

    the upper limit for example was

    x = 4 pi/3
    but
    u = x/8
    so the upper limit using u instead of x is
    u = (4 pi/3) /8
    or
    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

    so
    v at this upper limit is:
    v = sin (pi/6) = sin 30 degrees = 1/2

  • Calculus - Anti-Derivatives - ,

    Oh. Thank you, Damon, and Drwls. That really helped me out.

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