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sat_chen

find the integral of cot^3xcsc^3xdx

  • one year ago
  • one year ago

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  1. sat_chen
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    i got -csc^5x/5 + csc^3x/3 + C can someone pls tell me if im doing it right pls let me know thanks

    • one year ago
  2. mivanov
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    you want integral of cot^3(x)*csc^3(x) dx right?

    • one year ago
  3. escolas
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    is this (cotx)^3*(cscx)^3 dx?

    • one year ago
  4. sat_chen
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    yep

    • one year ago
  5. sat_chen
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    sorry about that should have made it more clear

    • one year ago
  6. mivanov
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    Take the integral: integral cot^3(x) csc^3(x) dx For the integrand cot^3(x) csc^3(x), use the trigonometric identity cot^2(x) = csc^2(x)-1: = integral cot(x) csc^3(x) (csc^2(x)-1) dx For the integrand cot(x) csc^3(x) (csc^2(x)-1), substitute u = csc(x) and du = -(cot(x) csc(x)) dx: = - integral u^2 (u^2-1) du Expanding the integrand u^2 (u^2-1) gives u^4-u^2: = - integral (u^4-u^2) du Integrate the sum term by term and factor out constants: = integral u^2 du- integral u^4 du The integral of u^4 is u^5/5: = integral u^2 du-u^5/5 The integral of u^2 is u^3/3: = u^3/3-u^5/5+constant Substitute back for u = csc(x): = (csc^3(x))/3-(csc^5(x))/5+constant Which is equal to: Answer: | | = -1/30 ((5 cos(2 x)+1) csc^5(x))+constant

    • one year ago
  7. mivanov
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    so it's just a simple substitution

    • one year ago
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