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anonymous

  • 5 years ago

o.k. intellectuals please sole for me the integral sin x^3 . cos x^3dx I know when both exp are odd ou can use either sub or convert method for smaller exp is easier I quess

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  1. anonymous
    • 5 years ago
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    this is a little crazy lol. hold on. thinking.

  2. anonymous
    • 5 years ago
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    ok well we can write it in this form. you can simplify the sin(x^3)cos(x^3) --> 0.5sin(2x^3)

  3. anonymous
    • 5 years ago
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    that's right, as 2 sinx cosx = sin(2x)

  4. anonymous
    • 5 years ago
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    now i m thinking what's the best way to get the integral

  5. anonymous
    • 5 years ago
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    the answer is 1/6sin (x)^2 cos (x)^4 - 1/12 cos (x0^4

  6. anonymous
    • 5 years ago
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    I don't know hoa to derive at that answer with the same exponents

  7. anonymous
    • 5 years ago
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    use substitution method from here. so you can do u = 2x^3 so du/dx = 6x^2 so you will get integrate (1/6x^2)sin(u) du and then from here we can write 6x^2 as some function of u itself. so like if u = 2x^3 and somethin along those lines

  8. anonymous
    • 5 years ago
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    Try Wolfram Alpha (wolframalpha.com)

  9. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=integral%20sin%28x%29^3*cos%28x%29^3

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