$\int \dfrac{1-sinx}{cos^2x}$ $dx$

The correct answer is $tanx$ - $secx$ + $C$

$\int \dfrac{1-sinx}{cos^2x}$ $dx$


$\int \left( \dfrac{1}{cos^2x} - \dfrac{sinx}{cos^2x} \right)$ $dx$


$\int \dfrac{1}{cos^2x}$ $dx$ - $\int \dfrac{sinx}{cos^2x}$ $dx$


$\int sec^2x$ $dx$ - $\int secx tanx$ $dx$


$tanx$ - $secx$ + $C$