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Math Help - Help with a trig integral

  1. #1
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    Mar 2010
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    Help with a trig integral

    OK I'm studying for a calculus exam and I ran across a problem where I thought I did everything right, but I can't know for sure because it's an even problem and I don't have an answer.

    Problem:

    \int cos^5 (x) dx =

    What I did:

    \int cos^5 (x) dx =

    Splitting into squares:
    \int [cos^2 (x)]^2 cos (x) dx

    Using cos^2 + sin^2 = 1:
    \int [1 - sin^2 (x)]^2 cos (x) dx

    Substituting:
    u = \sin (x)
    du = \cos (x) dx

    \int (1 - u^2)^2 dx

    Foiling:
    \int (1 - 2u^2 - u^4) dx

    Integrating:
    u - \frac{2}{3}u^3 - \frac{1}{5}u^5

    Unsubstituting:
    sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x)

    Answer:

    sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) + C

    I typed the problem into Wolfram Alpha and it gave me this:

     \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x) + C

    This looks alot different than my answer, and they did it using something called a 'reduction formula', which seems like something I would never remember. Is my answer wrong, or did they just do it differently? Or is Wolfram Alpha not the best place to be checking my answers?

    Thanks
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  2. #2
    Super Member General's Avatar
    Joined
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    Kuwait
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    Quote Originally Posted by bobhudy View Post
    OK I'm studying for a calculus exam and I ran across a problem where I thought I did everything right, but I can't know for sure because it's an even problem and I don't have an answer.

    Problem:

    \int cos^5 (x) dx =

    What I did:

    \int cos^5 (x) dx =

    Splitting into squares:
    \int [cos^2 (x)]^2 cos (x) dx

    Using cos^2 + sin^2 = 1:
    \int [1 - sin^2 (x)]^2 cos (x) dx

    Substituting:
    u = \sin (x)
    du = \cos (x) dx

    \int (1 - u^2)^2 dx

    Foiling:
    \int (1 - 2u^2 - u^4) dx

    Integrating:
    u - \frac{2}{3}u^3 - \frac{1}{5}u^5

    Unsubstituting:
    sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x)

    Answer:

    sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) + C

    I typed the problem into Wolfram Alpha and it gave me this:

     \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x) + C

    This looks alot different than my answer, and they did it using something called a 'reduction formula', which seems like something I would never remember. Is my answer wrong, or did they just do it differently? Or is Wolfram Alpha not the best place to be checking my answers?

    Thanks
    You did not make any mistake.
    Since: sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) \, =\, \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x).
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  3. #3
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    Joined
    Mar 2010
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    Quote Originally Posted by General View Post
    You did not make any mistake.
    Since: sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) \, =\, \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x).
    thanks. i now realise i shouldve just subbed in a number and seen if theyre the same -_-
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