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appleduardo
Group Title
hello, could somebody help me to solve the following integral? > sec^2 (ax)dx
 one year ago
 one year ago
appleduardo Group Title
hello, could somebody help me to solve the following integral? > sec^2 (ax)dx
 one year ago
 one year ago

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appleduardo Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{}^{}\sec ^{2}axdx\]
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.1
I know \[\sec^2 =1+\tan^2a\] but I don't know how to solve it
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
use u substitution where u = ax, so du = a*dx > dx = du/a
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
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

appleduardo Group TitleBest ResponseYou've already chosen the best response.1
so if do the substitution i'll get this >\[\frac{ 1 }{ a }\int\limits_{}^{}1+\tan^2ax\]
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.1
is that correct?
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
you could do that, but it would make things much tougher
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.1
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

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
good, assuming by tg, you mean tan
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
or tangent
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.1
yeep :D thank you so much! :D
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yw
 one year ago
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