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

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

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

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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1also 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
 one year ago
Best ResponseYou've already chosen the best response.1so if do the substitution i'll get this >\[\frac{ 1 }{ a }\int\limits_{}^{}1+\tan^2ax\]

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

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

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