Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

is that correct?

Answer 7

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

Answer 8

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?

Answer 9

good, assuming by tg, you mean tan

Answer 10

or tangent

Answer 11

yeep :D thank you so much! :D

Answer 12

yw