Question

Evaluate the Integral: \(\int {\frac{{1 - \sin x}}{{\cos x}}} dx\)

Short answer

To evaluate the integral, we usethe following identities:\(\sec x = \frac{1}{{\cos x}}\)and\(\tan x = \frac{{\sin x}}{{\cos x}}\)

Followed by integration formulas

\( = \int {\frac{{1 - \sin x}}{{\cos x}}} dx = \ln |\sec x + \tan x| - \ln |\sec x| + c\)

Step-by-step solution

Separation of Denominator

Let\(I = \int {\frac{{1 - \sin x}}{{\cos x}}} dx\)

First,we willseparate the denominator

\(I = \int {(\frac{1}{{\cos x}}} - \frac{{\sin x}}{{\cos x}})dx\)

Splitting of Integrals

Now,Split this integralas:

\(I = \int {\frac{1}{{\cos x}}} dx - \int {\frac{{\sin x}}{{\cos x}}} dx\)

Using General Trigonometric Identities

As\(\sec x = \frac{1}{{\cos x}}\)and\(\tan x = \frac{{\sin x}}{{\cos x}}\)

\(I = \int {\sec xdx - \int {\tan xdx} } \)

\(I = \ln |\sec x + \tan x| - \ln |\sec x| + c\)

Hencethe value ofthe integral:\(\int {\frac{{1 - \sin x}}{{\cos x}}} dx\)is

\( = \ln |\sec x + \tan x| - \ln |\sec x| + c\)