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heydayana

  • 2 years ago

evaluate the indefinite integral. problem below

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  1. heydayana
    • 2 years ago
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    \[\int\limits ( \Theta - \cos ( 1-\Theta) d \Theta \]

  2. artix_17
    • 2 years ago
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    1-sin(1-theta)/-1+C =1 +sin(1-theta)+c

  3. RadEn
    • 2 years ago
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    use integral subs, let u=(1-Θ)

  4. artix_17
    • 2 years ago
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    oops its I did a mistake there

  5. artix_17
    • 2 years ago
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    it should be \[\theta ^{2}/2 + \sin(1-\theta)+\]

  6. heydayana
    • 2 years ago
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    if i use subs what do i do after i get du=-1d\[\Theta \] ?

  7. RadEn
    • 2 years ago
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    well, i want explain it step by step u = 1-Θ -----> Θ=1-u, right ?

  8. heydayana
    • 2 years ago
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    yes

  9. RadEn
    • 2 years ago
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    next, u have got u = 1-Θ du = -dΘ or dΘ = - du, right ?

  10. heydayana
    • 2 years ago
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    yesss

  11. RadEn
    • 2 years ago
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    next, we substitute of them to the original problem : ∫(Θ−cos(1−Θ))dΘ = ∫[(1-u) - cosu] (-du) = ∫[(u-1) + cosu] du, agree ???

  12. heydayana
    • 2 years ago
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    yes

  13. RadEn
    • 2 years ago
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    ok, just integral all one by one, int u du= ... int -1du= .... int cosu du= .... what u get ????

  14. heydayana
    • 2 years ago
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    1/2 u^2 -u + sin u + c ?

  15. RadEn
    • 2 years ago
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    yes, correct now, the last step u must substitute back that u = 1-Θ

  16. heydayana
    • 2 years ago
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    i end up getting -1/2 -1/2theta^2 + sin (1-theta) is the -1/2 suppose to be there?

  17. RadEn
    • 2 years ago
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    from ur answer : 1/2 u^2 -u + sin u + c , just change u=1-Θ, gives 1/2(1-Θ)^2 -(1-Θ) + sin(1-Θ) + c i think enough be the answer, but if u want simplify it might too

  18. heydayana
    • 2 years ago
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    whats the final answer if you simplify?

  19. RadEn
    • 2 years ago
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    1/2(1-Θ)^2 -(1-Θ) + sin(1-Θ) + c = 1/2 (1-2Θ + Θ^2) -1 + Θ + sin(1-Θ) + c = 1/2 - Θ +1/2*Θ^2 -1 + Θ + sin(1-Θ) + c = -1/2 +1/2*Θ^2 + sin(1-Θ) + c = -1/2 (1-Θ^2) + sin(1-Θ) + c

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