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bettyboop8904

  • 3 years ago

can someone help me with integrals? equation below

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  1. bettyboop8904
    • 3 years ago
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    \[\int\limits_{}^{}tsec ^{2}2t\]

  2. khoala4pham
    • 3 years ago
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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.

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

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

  5. khoala4pham
    • 3 years ago
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    one half that...chain rule!

  6. bettyboop8904
    • 3 years ago
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    so it would be 2tan(2t)?

  7. bettyboop8904
    • 3 years ago
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    = (

  8. khoala4pham
    • 3 years ago
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    no no. 1/2 * tan2t

  9. bettyboop8904
    • 3 years ago
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    is that a law? = (

  10. khoala4pham
    • 3 years ago
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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

  11. bettyboop8904
    • 3 years ago
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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) ?

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

  13. khoala4pham
    • 3 years ago
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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)

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

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

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

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