Question

Write a value of $\int \frac{1-\sin x}{{\cos }^{2}x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{1-\sin x}{{\cos }^{2}x}\right)dx \\ =\int \left(\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}\right)dx \\ \int \left(\frac{1}{{\cos }^{2}x}-\frac{\sin x}{\cos x}\times \frac{1}{\cos x}\right)dx \\ =\int \left({\sec }^{2}x-\sec x\tan x\right)dx \\ =\tan x-\sec x+C \end{gathered}$$