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



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appleduardo
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\[\int\limits_{}^{}\sec ^{2}axdx\]

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

jim_thompson5910
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use u substitution where u = ax, so du = a*dx > dx = du/a

jim_thompson5910
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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

appleduardo
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so if do the substitution i'll get this >\[\frac{ 1 }{ a }\int\limits_{}^{}1+\tan^2ax\]

appleduardo
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is that correct?

jim_thompson5910
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you could do that, but it would make things much tougher

appleduardo
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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?

jim_thompson5910
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good, assuming by tg, you mean tan

jim_thompson5910
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or tangent

appleduardo
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yeep :D thank you so much! :D
