appleduardo
  • appleduardo
hello, could somebody help me to solve the following integral? --> sec^2 (ax)dx
Mathematics
jamiebookeater
  • jamiebookeater
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appleduardo
  • appleduardo
\[\int\limits_{}^{}\sec ^{2}axdx\]
appleduardo
  • appleduardo
I know \[\sec^2 =1+\tan^2a\] but I don't know how to solve it
jim_thompson5910
  • jim_thompson5910
use u substitution where u = ax, so du = a*dx ---> dx = du/a

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jim_thompson5910
  • jim_thompson5910
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
appleduardo
  • appleduardo
so if do the substitution i'll get this -->\[\frac{ 1 }{ a }\int\limits_{}^{}1+\tan^2ax\]
appleduardo
  • appleduardo
is that correct?
jim_thompson5910
  • jim_thompson5910
you could do that, but it would make things much tougher
appleduardo
  • appleduardo
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?
jim_thompson5910
  • jim_thompson5910
good, assuming by tg, you mean tan
jim_thompson5910
  • jim_thompson5910
or tangent
appleduardo
  • appleduardo
yeep :D thank you so much! :D
jim_thompson5910
  • jim_thompson5910
yw

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