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appleduardo

  • 3 years ago

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

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  1. appleduardo
    • 3 years ago
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    \[\int\limits_{}^{}\sec ^{2}axdx\]

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

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

  4. jim_thompson5910
    • 3 years ago
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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

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

  6. appleduardo
    • 3 years ago
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    is that correct?

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

  8. appleduardo
    • 3 years ago
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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?

  9. jim_thompson5910
    • 3 years ago
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    good, assuming by tg, you mean tan

  10. jim_thompson5910
    • 3 years ago
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    or tangent

  11. appleduardo
    • 3 years ago
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    yeep :D thank you so much! :D

  12. jim_thompson5910
    • 3 years ago
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    yw

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