[tex]\displaystyle \sf \int \frac{sen(x)}{cos^2(x)}dx\to -\int \frac{-sen(x)}{cos^2(x)}dx \\\\\\ \text{fa\c camos} : \\\\ u = cos(x) \to du= -sen(x)dx \\\\ da{\'i}}: \\\\\ -\int \frac{du}{u^2} \to - \int u^{-2}du \to -\frac{u^{-2+1}}{-2+1} = u^{-1}+C \\\\\\ u^{-1}+C = \frac{1}{cos(x)}+C \\\\\\ \frac{1}{cos(x)}+C = sec(x)+C \\\\\\\ Portanto : \\\\ \boxed{\sf \ \int\frac{sen(x)}{cos^2(x)}dx =sec(x)+C\ }\checkmark[/tex]

Resposta:

∫ sen(x)/cos²(x) dx

Fazendo u= cos(x)   ==>du= -sen(x) dx

∫ sen(x)/u² du/(-sen(x) )

-∫ 1/u² du )

-∫ u⁻² du )

= - u⁻¹/(-1) +c

= 1/u  +c

Como u= cos(x)

=1/cos(x)+c

=sec(x) +c