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anonymous

  • one year ago

I'm confused a little... How does the integral of secxtanx = secx ?

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  1. anonymous
    • one year ago
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    it doesnt ._.

  2. anonymous
    • one year ago
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    Really? XD lol Then what is is?

  3. anonymous
    • one year ago
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    it*

  4. anonymous
    • one year ago
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    sec(theta)=\[\frac{ 1 }{ \cos*theta}\]

  5. anonymous
    • one year ago
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    Right

  6. jim_thompson5910
    • one year ago
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    if y = sec(x), then dy/dx = sec(x)*tan(x) take this in reverse to get \[\Large \int(\sec(x)\tan(x))dx = \sec(x) + C\]

  7. anonymous
    • one year ago
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    Ohhh right! That makes sense!!

  8. jim_thompson5910
    • one year ago
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    if you had no idea that dy/dx = sec(x)*tan(x) then you can convert sec(x)*tan(x) into (1/cos)*(sin/cos) which becomes sin/(cos^2) from there you use u-sub u = cos(x) du = -sin(x)dx

  9. UsukiDoll
    • one year ago
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    that's one of the standard derivative definitions... The derivative of secx is secxtanx the antiderivative of secxtanx is secx.

  10. anonymous
    • one year ago
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    Yeah I would've had to go through it with u-sub... i didn't know that dy/dx of secx = secxtanx

  11. anonymous
    • one year ago
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    Right! Thanks you guys! That really helped!

  12. jim_thompson5910
    • one year ago
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    you're welcome

  13. anonymous
    • one year ago
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    :)

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