Question

Calcule o valor da integral indefinida: tgx dx

Answer 1

$\int\ {tanx} \, dx$

$\int\ {\frac{sinx}{cosx}} \, dx$

$u=cosx$

$du=-sinxdx$

$dx=\frac{du}{-sinx}$

$\int\ {\frac{sinx}{u}} \, \frac{du}{-sinx}$

$\int\ {\frac{1}{-u}} \, du$

$-\int\ {\frac{1}{u}} \, du$

$-\int\ {u^{-1}}} \, du$

$-ln(u)+C$

$-ln(cosx)+C$