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

1. appleduardo Group Title

$\int\limits_{}^{}\sec ^{2}axdx$

2. appleduardo Group Title

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

3. jim_thompson5910 Group Title

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

4. jim_thompson5910 Group Title

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

5. appleduardo Group Title

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

6. appleduardo Group Title

is that correct?

7. jim_thompson5910 Group Title

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

8. appleduardo Group Title

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?

9. jim_thompson5910 Group Title

good, assuming by tg, you mean tan

10. jim_thompson5910 Group Title

or tangent

11. appleduardo Group Title

yeep :D thank you so much! :D

12. jim_thompson5910 Group Title

yw