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bettyboop8904

can someone help me with integrals? equation below

  • one year ago
  • one year ago

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  1. bettyboop8904
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    \[\int\limits_{}^{}tsec ^{2}2t\]

    • one year ago
  2. khoala4pham
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    like your last problem, this is another integration by parts. How do we know? because sec^2 is ordinarily integrable but it's multiplied by t. Here, we let u = t, and dv = sec^2 (t). du = dt and v = tan(t) with this information, you should be able to solve.

    • one year ago
  3. bettyboop8904
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    lol i literally just figured it out right when you posted that lol ty though = )

    • one year ago
  4. bettyboop8904
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    one question though dv=sec^2 (2t) would be v=tan(2t) right?

    • one year ago
  5. khoala4pham
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    one half that...chain rule!

    • one year ago
  6. bettyboop8904
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    so it would be 2tan(2t)?

    • one year ago
  7. bettyboop8904
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    = (

    • one year ago
  8. khoala4pham
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    no no. 1/2 * tan2t

    • one year ago
  9. bettyboop8904
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    is that a law? = (

    • one year ago
  10. khoala4pham
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    ? well, the chain rule is not a "law" either you believe in it or you don't. If you don't then most mathematicians just won't work with you. :P

    • one year ago
  11. bettyboop8904
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    so wait everytime you have a trig sign with f(x) and you take the integral its that trig sign with f(x) multiplied by 1/f(x) ?

    • one year ago
  12. bettyboop8904
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    i'm sorry maybe i only remember the chain rule for derivatives... not integrals...

    • one year ago
  13. khoala4pham
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    Well I mean it's a simple consequence of u substitution is what I mean. Suppose you wanted to integrate cos2t. you technically only know the formula for the integral of cost. But by being clever, you let u = 2t. du = 2dt. thus dt = (1/2)du. AGAIN I MUST REMIND YOU THESE ARE NOT FRACTIONS but they work out that way for these general cases. So (integral)cos2tdt = (integral) (1/2) cosu du. Why? because originally, we integrated with respect to dt. So we must change everything that is in terms of to to everything in terms of u. That is why this integral evaluates to 1/2 tan(2t)

    • one year ago
  14. bettyboop8904
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    ok thank you i'm sorry i suck at this = (

    • one year ago
  15. khoala4pham
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    Hey no problem. It's always good to learn. We each have our own pace.

    • one year ago
  16. bettyboop8904
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    mine seems to be slower than a snails' = (

    • one year ago
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