Question:
\( \int \frac{dx}{\sec x + \tan x} \) is equal to
\( \int \frac{dx}{\sec x + \tan x} \) is equal to
Show Hint
Always look for "derivative-function" pairs in trigonometric integrals; here \( \cos x \) is the derivative of \( \sin x \).
Rationalizing the denominator by multiplying by \( (\sec x - \tan x) \) is another valid method.
Rationalizing the denominator by multiplying by \( (\sec x - \tan x) \) is another valid method.
Updated On: Sep 10, 2026
- \( \log | \sec x + \tan x | + C \)
- \( \log | \sec x - \tan x | + C \)
- \( \log | 1 + \cos x | + C \)
- \( \log | 1 + \sin x | + C \)
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The Correct Option is D
Solution and Explanation
Concept:
• Integration of trigonometric functions.
• Substitution method: Replacing part of the integrand to simplify the integration process.
• Trigonometric identity: \( \sec x + \tan x = \frac{1 + \sin x}{\cos x} \).
Step 1: Simplify the integrand using basic trigonometric identities
Convert the secant and tangent functions into sine and cosine:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]
The integral becomes:
\[ I = \int \frac{1}{\frac{1 + \sin x}{\cos x}} dx = \int \frac{\cos x}{1 + \sin x} dx \]
Step 2: Perform a substitution
Let \( u = 1 + \sin x \).
Then, differentiating with respect to \( x \):
\[ du = \cos x \, dx \]
Step 3: Evaluate the integral
Substitute \( u \) and \( du \) into the integral:
\[ I = \int \frac{1}{u} du \]
\[ I = \log |u| + C \]
Substituting back for \( u \):
\[ I = \log |1 + \sin x| + C \]
This matches option (D).
• Integration of trigonometric functions.
• Substitution method: Replacing part of the integrand to simplify the integration process.
• Trigonometric identity: \( \sec x + \tan x = \frac{1 + \sin x}{\cos x} \).
Step 1: Simplify the integrand using basic trigonometric identities
Convert the secant and tangent functions into sine and cosine:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]
The integral becomes:
\[ I = \int \frac{1}{\frac{1 + \sin x}{\cos x}} dx = \int \frac{\cos x}{1 + \sin x} dx \]
Step 2: Perform a substitution
Let \( u = 1 + \sin x \).
Then, differentiating with respect to \( x \):
\[ du = \cos x \, dx \]
Step 3: Evaluate the integral
Substitute \( u \) and \( du \) into the integral:
\[ I = \int \frac{1}{u} du \]
\[ I = \log |u| + C \]
Substituting back for \( u \):
\[ I = \log |1 + \sin x| + C \]
This matches option (D).
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