Question

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

• A. $\tan x-\frac{1}{\cos x}+c$
• B. $\cot x-\frac{1}{\cos x}+c$
• C. $\tan x+\frac{1}{\cos x}+c$
• D. None of these

Answer 1

The correct option is A $\tan x-\frac{1}{\cos x}+c$

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

$={\int }^{}\left(\frac{1}{{\cos x}^2}-\frac{\sin x}{{\cos }^{2}x}\right)dx$

$={\int }^{}\left({\sec }^{2}x-\sec x\tan x\right)dx$

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

$=\tan x-\frac{1}{\cos x}+c$