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appleduardo

hello, could somebody help me to solve the following integral? --> sec^2 (ax)dx

  • one year ago
  • one year ago

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  1. appleduardo
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    \[\int\limits_{}^{}\sec ^{2}axdx\]

    • one year ago
  2. appleduardo
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    I know \[\sec^2 =1+\tan^2a\] but I don't know how to solve it

    • one year ago
  3. jim_thompson5910
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    use u substitution where u = ax, so du = a*dx ---> dx = du/a

    • one year ago
  4. jim_thompson5910
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    also use this if y = tan(x), then y ' = sec^2(x) so if f(x) = sec^2(x), then the integral of f(x) is tan(x) + C

    • one year ago
  5. appleduardo
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    so if do the substitution i'll get this -->\[\frac{ 1 }{ a }\int\limits_{}^{}1+\tan^2ax\]

    • one year ago
  6. appleduardo
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    is that correct?

    • one year ago
  7. jim_thompson5910
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    you could do that, but it would make things much tougher

    • one year ago
  8. appleduardo
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    so uhm I have to make use of \[\int\limits_{}^{}\sec^2 u du = tg(u)+c\] and i'll get:\[\frac{ 1 }{ a }tg(ax)+c\] ?? is that correct?

    • one year ago
  9. jim_thompson5910
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    good, assuming by tg, you mean tan

    • one year ago
  10. jim_thompson5910
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    or tangent

    • one year ago
  11. appleduardo
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    yeep :D thank you so much! :D

    • one year ago
  12. jim_thompson5910
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    yw

    • one year ago
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