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ksaimouli

  • 2 years ago

integratebb

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  1. ksaimouli
    • 2 years ago
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    3siny sqtcosy dy

  2. ksaimouli
    • 2 years ago
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    \[\int\limits_{}^{}3\sin y \sqrt{cosy} \]

  3. Jemurray3
    • 2 years ago
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    substitute u = cos(y), du = -sin(y) dy

  4. ksaimouli
    • 2 years ago
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    yup i did i am getting \[-3\sqrt{cosy}\]

  5. Jemurray3
    • 2 years ago
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    but cos(y) is u.

  6. Jemurray3
    • 2 years ago
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    \[ \int -3\sqrt{u} du \]is a pretty easy integral.

  7. ksaimouli
    • 2 years ago
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    |dw:1357088383865:dw|

  8. ksaimouli
    • 2 years ago
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    see i am getting <-3 {sqrtcosy}/>

  9. ksaimouli
    • 2 years ago
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    \[-3 {\sqrt cosy}\]

  10. Jemurray3
    • 2 years ago
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    I reiterate that cos(y) is just u.

  11. Jemurray3
    • 2 years ago
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    so that is -3 sqrt(u)

  12. ksaimouli
    • 2 years ago
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    yes thats what i did -3 sqrt(u) and before we said u=cosx so i rewrote it back

  13. ksaimouli
    • 2 years ago
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    is that right answer for this question 3siny sqtcosy dy

  14. Jemurray3
    • 2 years ago
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    No. After your substitution the integral becomes \[ \int -3 \sqrt{u} du\] you need to integrate this, which you should be able to do in two seconds. Then, you can replace all the u's with cos(y)'s and then you'll be done.

  15. MrDoe
    • 2 years ago
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    jemurray3 is correct, just think of the square root as u^1/2 i think thats what your having trouble with

  16. ksaimouli
    • 2 years ago
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    -2u^(3/2)

  17. ksaimouli
    • 2 years ago
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    thx

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