Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculus II - Trigonometric Integrals

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate integral( sin^3(x) cos^5(x) ) dx


    2. Relevant equations

    sin^2(x) + cos^2(x) = 1
    integral x^n dx = x^(n+1)/(n+1) + c
    d/dx cos(x) = -sin(x)
    a^n*a^m=a^(n+m)

    3. The attempt at a solution

    I got -cos^6(x)/6+cos^8(x)/8+c
    Apparently I did something wrong
    SEE ATTACHMENT

    Thank you for any assistance! I really don't see what I'm doing wrong. I believe that the method I chose to evaluate this integral is overly simple but mathematically correct and I don't know what I'm doing wrong.
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2011 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your answer is correct; why do you think it is wrong? Here it is in Maple 9.5:
    Int( sin(x)^3 *cos(x)^5,x);J1:=value(%);
    /
    |
    | sin(x)^3 cos(x)^5 dx
    |
    /



    J1 := -1/8 sin(x)^2 cos(x)^6 - 1/24 cos(x)^6 <---indef. integral
    simplify(diff(J1,x));

    sin(x)^3 cos(x)^5 <---- derivative is OK
    expand(subs(sin(x)^2=1-cos(x)^2,J1));

    -1/6 cos(x)^6 + 1/8 cos(x)^8
    That's your answer.

    RGV
     
    Last edited: Jul 30, 2011
  4. Jul 30, 2011 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi GreenPrint! :smile:

    (try using the X2 icon just above the Reply box :wink:)

    Looks ok to me. :confused:

    (what's the official answer?)
     
  5. Jul 30, 2011 #4
    Ahhh wolfram alpha is giving me answers that don't match mine and so I think that I am wrong...

    wolfram alpha
    http://www.wolframalpha.com/input/?i=integral%28+sin^3%28x%29+cos^5%28x%29+%29+dx
    (-72cos(2x)-12cos(4x)+8cos(6x)+3cos(8x))/3072 + constant

    and my answer
    -cos^6(x)/6+cos^8(x)/8+c

    they don't match which is fine but they don't appear to be equivalent to each other

    lol i should start using that button, thanks
     
  6. Jul 30, 2011 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi GreenPrint! :wink:
    They're probably the same …

    try rewriting the first one using 1 + cos(2x) = 2cos2x etc …

    what do you get? :smile:
     
  7. Jul 30, 2011 #6
    -cos^6(5)/6+cos^8(5)/8 is about -8.15876413*10^-5
    (-72cos(2*5)-12cos(4*5)+8cos(6*5)+3cos(8*5))/3072 is about .0178220582

    umm hmmm i don't think i have seen that formula in over 3 years... uh lets see

    I don't know I don't think I remember enough of the pre calc formulas to do that >_>
     
    Last edited: Jul 30, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook