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Trigonomic integrals

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data

    do my professor did this in class, and it doesn't make sense to me
    ∫cos^5(x) sin^4(x) dx
    ∫cos^4(x) sin^4(x) cos(x) dx
    ∫(1-sin^2(x))^2 sin^4(x) cos(x) dx
    ∫(sin^4(x) -2sin^6(x) +sin^8(x))cos(x) dx

    so the above I get but when my professor integrated I became lost, this is his solution

    1/5 sin^5(x) - 2/7 sin^7(x) +1/9 sin^9(x)

    shouldn't those sin be cos? And where did the cos on the outside go?
  2. jcsd
  3. Mar 7, 2012 #2


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    hi ex81! :smile:

    (try using the X2 button just above the Reply box :wink:)
    eg (sin9x)' = 9sin8x(sinx)' = 9sin8xcosx :wink:

    (from the chain rule … (f(g(x))' = f'(g(x))g'(x))
  4. Mar 7, 2012 #3


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    Another way to think about it is that you are making the substitution y= sin(x) so that dy= cos(x) dx. The integral becomes
    [tex]\int (y^4- 2y^6+ y^8)dy[/tex]

    integrate that and replace y with sin(x).
  5. Mar 11, 2012 #4
    I think I know where you are going.

    Lets try another one :D

    ∫cos^5(x)/(sin(x))^1/2 dx
    so the next step would be to split a cos off of that
    ∫cos^4(x) cos(x) /(sin(x))^1/2 dx
    ∫(1-sin^2(x))^2 (cos(x) /(sin(x))^1/2 dx
    ∫ (1-2sin^2(x) + sin^4(x)) cos(x) /(sin(x))^1/2 dx

    This is where I get stuck, and it seems more complicated than it needs to be but I am not sure where to go
  6. Mar 11, 2012 #5


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    then eg the ∫ sin4x cosx / sin1/2x dx

    = ∫ sin3.5x cosx dx

    = … ? :smile:
  7. Mar 11, 2012 #6
    Well, that is definitely different than what I was trying to do. Looks way more simple :)

    hehe, go figure I wouldn't look at it from a POV of long division

    ∫(1/(sin(x))^1/2 -2sin^3/2 (x) + sin^7/2 (x) )cos (x)

    so that would be 2 sin^1/2 (x) -4/5 sin^5/2 (x) + 2/9 sin^9/2 (x) +C

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