Qual é a integral de int 1 / (secx + tanx) dx ∫1secx+tanxdx?
Responda:
int 1/(sec x+tan x)d x=l n(1+sin x) + C∫1secx+tanxdx=ln(1+sinx)+C
Explicação:
int 1/(sec x+tan x)d x=(d x)/(1/cos x+sin x/cos x)∫1secx+tanxdx=dx1cosx+sinxcosx
int (cos x d x)/(1+sin x)" "1+sin x=u" "cos x d x=d u∫cosxdx1+sinx 1+sinx=u cosxdx=du
int (d u)/u∫duu
int 1/(sec x+tan x)d x=l n u +C∫1secx+tanxdx=lnu+C
int 1/(sec x+tan x)d x=l n(1+sin x) + C∫1secx+tanxdx=ln(1+sinx)+C