Question

Evaluate: $ \int {\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)} dx $ ?

Answer 1

Hint: In the given question, we are required to integrate the given function with respect to the variable x. One must have a strong grip order trigonometry and integration in order to solve such questions correctly. We should know the integrals of some basic functions like sine, cosine, secant, tangent, etc. in order to solve such kinds of questions.

Complete step-by-step answer:
The given question requires us to integrate a rational trigonometric function $ \left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right) $ in variable x whose numerator is $ \left( {1 - \sin x} \right) $ and denominator is $ \cos x $ . So, we first simplify the rational trigonometric function given to us and then integrate the function directly using the results of integration of basic trigonometric functions.
So, we have, $ \int {\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)} dx $
Separating the two fractions by splitting up the denominator underneath both the terms in numerator, we get,
 $ \Rightarrow \int {\left( {\sec x - \tan x} \right)} dx $
Now, we know the integrals of the trigonometric functions $ \sec x $ and $ \tan x $ .
So, we have the integration of $ \sec x $ with respect to x as $ \ln \left| {\sec x + \tan x} \right| $ and the integration of $ \tan x $ with respect to x as $ \ln \left| {\sec x} \right| $ .
So, we get the integral as,
 $ \Rightarrow \ln \left| {\sec x + \tan x} \right| - \ln \left| {\sec x} \right| + c $ , where c is any arbitrary constant.
Now, we have to simplify the expression so as to match one of the options given to us in the question itself.
So, we can condense the two logarithms together with the help of law of logarithms $ \ln \left( a \right) - \ln \left( b \right) = \ln \left( {\dfrac{a}{b}} \right) $ . So, we get,
 $ \Rightarrow \ln \left| {\dfrac{{\sec x + \tan x}}{{\sec x}}} \right| + c $
Now, we express all the trigonometric ratios in terms of sine and cosine in order to simplify the expression and match the options given in the question. So, we get,
 $ \Rightarrow \ln \left| {\dfrac{{\dfrac{{1 + \sin x}}{{\cos x}}}}{{\dfrac{1}{{\cos x}}}}} \right| + c $
So, cancelling the common factors, we get,
 $ \Rightarrow \ln \left| {1 + \sin x} \right| + c $
So, the value of the integral $ \int {\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)} dx $ is equal to $ \ln \left| {1 + \sin x} \right| + c $ , where c is any arbitrary constant.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant.