Question

integral de dx/ cos4x

Answer 1

∫dx/ cos(4x)
u= 4x; du= 4dx
∫dx/ cos(4x) = 1/4 ∫4dx/ cos(4x)

=1/4 ∫du/ cos(u) = 1/4 ∫ sec u du = 1/4 ∫ sec u (sec u+tanu)/(sec u+tanu) du
= 1/4 ∫ sec² u + secu tanu/(sec u+tanu) du
a= sec u + tan u
da= (sec²u + sec u tan u) du

Luego
1/4 ∫ sec² u + secu tanu/(sec u+tanu) du = 1/4∫da/a = 1/4 ln lal + c
= 1/4 lnlsec u + tan ul + c = 1/4 ln lsec(4x)+tan(4x)l + c