$∫\frac{1-sinx}{cos^2x}dx$ is equal to :

$tanx-secx+C$ Where C is the constant of integration

Integral: $\int \frac{1-\sin x}{\cos^{2}x}\,dx$

Split terms:

$\int \frac{1}{\cos^{2}x}\,dx - \int \frac{\sin x}{\cos^{2}x}\,dx$

$=\int \sec^{2}x\,dx - \int \frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}\,dx$

$=\int \sec^{2}x\,dx - \int \tan x \sec x\,dx$

$=\tan x - \sec x + C$

Required answer = $\tan x - \sec x + C$