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ksaimouli

  • one year ago

integratebb

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  1. ksaimouli
    • one year ago
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    3siny sqtcosy dy

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

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

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

  5. Jemurray3
    • one year ago
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    but cos(y) is u.

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

  7. ksaimouli
    • one year ago
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    |dw:1357088383865:dw|

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

  9. ksaimouli
    • one year ago
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    \[-3 {\sqrt cosy}\]

  10. Jemurray3
    • one year ago
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    I reiterate that cos(y) is just u.

  11. Jemurray3
    • one year ago
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    so that is -3 sqrt(u)

  12. ksaimouli
    • one year 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
    • one year ago
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    is that right answer for this question 3siny sqtcosy dy

  14. Jemurray3
    • one year 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
    • one year 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
    • one year ago
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    -2u^(3/2)

  17. ksaimouli
    • one year ago
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    thx

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