Question

The integral of $\int \frac{1-\sin x}{{\cos }^{2}x}dx$ is

$\left(a\right)\tan x+\sec x+c\left(b\right)\tan x-\sec x+c$
$\left(a\right)\sec x-\tan x+c\left(b\right)\tan x-\ln \sec x+c$

Answer 1

$I=\int \frac{1-\sin x}{{\cos }^{2}x}dx$
$=\int \frac{1}{{\cos }^{2}x}dx-\int \frac{\sin x}{{\cos }^{2}x}dx$
$=\int {\sec }^{2}xdx-\int \tan x\sec xdx$
$=\tan x-\sec x+c$

Option $\left(b\right)$ is correct.