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

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

appleduardoBest 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_thompson5910Best 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_thompson5910Best 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

appleduardoBest 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

jim_thompson5910Best ResponseYou've already chosen the best response.1
you could do that, but it would make things much tougher
 one year ago

appleduardoBest 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_thompson5910Best ResponseYou've already chosen the best response.1
good, assuming by tg, you mean tan
 one year ago

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