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anonymous

  • one year ago

Need help! How does integral of sinx - cscx become -cosx?

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  1. anonymous
    • one year ago
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    |dw:1438832459561:dw|

  2. anonymous
    • one year ago
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    ...does anyone know whats going on?

  3. geerky42
    • one year ago
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    It doesn't.

  4. geerky42
    • one year ago
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    http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=integral%20of%20sin%20x%20-%20csc%20x

  5. anonymous
    • one year ago
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    Well.. thats the answer in the back of the book buddy

  6. geerky42
    • one year ago
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    Well, textbooks aren't perfect.

  7. mathstudent55
    • one year ago
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    |dw:1438832964732:dw|

  8. anonymous
    • one year ago
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    Okay, so let me just write down the original integral problem

  9. anonymous
    • one year ago
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    |dw:1438833089151:dw|

  10. anonymous
    • one year ago
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    So that was the original problem... and know Im trying to figure out how to get to the answer... which apparently is: sec(x)+c

  11. anonymous
    • one year ago
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    *now

  12. anonymous
    • one year ago
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    wait, whyd you delete it?!

  13. geerky42
    • one year ago
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    HINT: Use u-substitution here. \[\int \dfrac{\sin x}{\cos^2 x} ~\mathrm dx\]

  14. anonymous
    • one year ago
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    We havent learned that

  15. geerky42
    • one year ago
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    what do you know so far?

  16. anonymous
    • one year ago
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    We have never applied U substitution method so far in class

  17. geerky42
    • one year ago
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    What method did you learn?

  18. anonymous
    • one year ago
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    Just evaluating the integral normally

  19. geerky42
    • one year ago
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    Ok, do you use integral table or something?

  20. geerky42
    • one year ago
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    You do know that \(\dfrac{\sin x}{\cos^2x} = \tan x\sec x\)?

  21. anonymous
    • one year ago
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    wouldnt it be adding tanx + sec x

  22. anonymous
    • one year ago
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    Oh no, youre right

  23. geerky42
    • one year ago
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    So can you evaluate integral now?

  24. anonymous
    • one year ago
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    The answer is clear now

  25. anonymous
    • one year ago
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    thanks

  26. geerky42
    • one year ago
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    okay good.

  27. geerky42
    • one year ago
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    no problem

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