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wendo

  • 3 years ago

1. integral sin x/(cos x )^2= ? 2. integral (2-3t)^6 dt=?

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  1. dpaInc
    • 3 years ago
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    for the first one, u=cosx

  2. dpaInc
    • 3 years ago
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    second one u=2-3t

  3. lgbasallote
    • 3 years ago
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    this doesnt seem like a u-sub problem @dpaInc looks like that basic application of general power formula

  4. Romero
    • 3 years ago
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    @lgbasallote Not for the first one

  5. dpaInc
    • 3 years ago
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    not necessarily but i like to do it so I don't miss the derivative of 2-3t...

  6. lgbasallote
    • 3 years ago
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    me too...that formula is too confusing :/ kinda like doing limits to find derivative

  7. dpaInc
    • 3 years ago
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    and @Romero , yes there's more than one way to skin this cat.. for the first one, the integrand can be rewritten as secxtanx

  8. wendo
    • 3 years ago
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    from my caculation first one is 1/cosx+c second one is -(2-3t)^7/21 +c is that right

  9. lgbasallote
    • 3 years ago
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    i think there should be a negative on #1

  10. lgbasallote
    • 3 years ago
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    *think*

  11. campbell_st
    • 3 years ago
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    looks like a nice simple substitution to me... but I'm still waiting for the rooster eggs to roll off the roof

  12. lgbasallote
    • 3 years ago
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    #2 seems right though @campbell_st whatis it with you and eggs :p

  13. Romero
    • 3 years ago
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    Second one is right. You just wrote it weird so it looks like the ^7 is getting divided by 21

  14. wendo
    • 3 years ago
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    why #1 is negative?

  15. Romero
    • 3 years ago
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    derivative of cos will give you a negative so when you do the chain rule you will need to cancel that negative with another one.

  16. quarkine
    • 3 years ago
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    these are too basic...perhaps u need to revisit the integration methods... 1/cosx and ((2-3t)^7)/(-3*7)

  17. freckles
    • 3 years ago
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    \[\int\limits_{}^{}\frac{\sin(x)}{(\cos(x))^2} dx ...u=\cos(x) =>du=-\sin(x) dx ......\int\limits_{}^{}\frac{-du}{u^2}=-\int\limits_{}^{}u^{-2} du\] \[\int\limits_{}^{}(2-3t)^6 dt....u=2-3t=> du=-3 dt .....\int\limits_{}^{}u^6 \frac{du}{-3}\]

  18. .Sam.
    • 3 years ago
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    #2 \[\int\limits (2-3 t)^6 \, dt\] u==2-3t, du=-3dt \[-\frac{1}{3}\int\limits u^6 \, du\] \[-\frac{u^7}{21}+c\] \[-\frac{(2-3 t)^7}{21} +c\]

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