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anonymous

  • 4 years ago

Need help with another intergral. Never really done one like this before. tan^6(3y)sec^2(3y)dy

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  1. anonymous
    • 4 years ago
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    fo u know c++

  2. anonymous
    • 4 years ago
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    \[\int\limits_{?}^{?}\tan ^6\left( 3y \right)\sec ^2(3y)dy\]

  3. anonymous
    • 4 years ago
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    I think I need to do u=tan(3y)? Maybe not even close? ,

  4. anonymous
    • 4 years ago
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    or u=tanx? So du=sec^2x?

  5. anonymous
    • 4 years ago
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    Anyone have any clue?

  6. anonymous
    • 4 years ago
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    turing test ..do u knw c++?

  7. TuringTest
    • 4 years ago
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    yes, that will work, but it's\[u=\tan(3y)\]\[du=3\sec^2(3y)\]remember to use the chain rule all the way through @shiv no, sorry

  8. TuringTest
    • 4 years ago
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    sorry\[u=\tan(3y)\]\[du=3\sec^2(3y)dy\]can't forget the dy! very important

  9. anonymous
    • 4 years ago
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    so would it be \[1/3\int\limits_{?}^{?}\tan(u)^6dy?\]

  10. TuringTest
    • 4 years ago
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    no,\[u=\tan(3y)\]so after the sub there should be no tangent left

  11. anonymous
    • 4 years ago
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    oh... its \[1/3\int\limits_{?}^{?}u^6du\]

  12. TuringTest
    • 4 years ago
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    yep :D

  13. anonymous
    • 4 years ago
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    you are a life saver. Can I just ask you one more quick here

  14. TuringTest
    • 4 years ago
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    sure but I'm headed for dinner so gotta be quick

  15. anonymous
    • 4 years ago
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    ill try. \[\int\limits_{-1}^{-1/2}t^-2 \sin^2(1+1/2)dt sorry its t \to the -2 wont show up though\]

  16. anonymous
    • 4 years ago
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    Any direction helps. Do i have to set something u ,, then eventually change the limits of integration? I never know when to do that and feel this could be one..

  17. TuringTest
    • 4 years ago
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    \[\int t^{-2}\sin^2?dt\]sorry, what is the integrand exactly?

  18. anonymous
    • 4 years ago
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    the end is sin^2(1+1/t)

  19. anonymous
    • 4 years ago
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    that 2 was wrong. you got the rest right. evaluated from -1 to -.5

  20. TuringTest
    • 4 years ago
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    \[\int t^{-2}\sin^2(1+\frac1t)dt\]yes?

  21. anonymous
    • 4 years ago
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    yes

  22. anonymous
    • 4 years ago
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    evaluated from -1 to -1/2 if that changes anything

  23. TuringTest
    • 4 years ago
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    \[u=1+\frac1t\]you don't have to change the bounds if you turn it back to t at the end, but if you want to evaluate it in terms of u you need to do this:\[u_1=1+\frac1{-1}=0\]\[u_2=1+\frac1{-1/2}=-1\]so those are your new limits (if you want to evaluate in terms of u the differential is\[du=-t^{-2}dt\]so the integral is\[-\int_{0}^{-1}\sin udu=\int_{-1}^{0}\sin u du\]Ok gotta eat, good luck!

  24. anonymous
    • 4 years ago
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    Thanks a lot! I appreicate it

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