Finding the Indefinite Integral of \(\frac{1}{\sec x + \tan x}\)

We need to evaluate the integral \(\int \frac{dx}{\sec x+\tan x}\). This integral involves trigonometric functions in the denominator, which can often be simplified.

Simplifying the Integrand

A common technique for expressions involving \(\sec x + \tan x\) or \(\sec x - \tan x\) in the denominator is to multiply the numerator and the denominator by the conjugate. The conjugate of \(\sec x + \tan x\) is \(\sec x - \tan x\).

Let's multiply the integrand by \(\frac{\sec x - \tan x}{\sec x - \tan x}\):

\[ \frac{1}{\sec x + \tan x} \times \frac{\sec x - \tan x}{\sec x - \tan x} = \frac{\sec x - \tan x}{(\sec x + \tan x)(\sec x - \tan x)} \]

Now, we use the difference of squares formula, \( (a+b)(a-b) = a^2 - b^2 \). Here \(a = \sec x\) and \(b = \tan x\). So, the denominator becomes \(\sec^2 x - \tan^2 x\).

Recall the fundamental trigonometric identity: \(\sec^2 x - \tan^2 x = 1\).

Therefore, the integrand simplifies to:

\[ \frac{\sec x - \tan x}{\sec^2 x - \tan^2 x} = \frac{\sec x - \tan x}{1} = \sec x - \tan x \]

Evaluating the Integral

Now, the integral becomes much simpler:

\[ \int \frac{dx}{\sec x+\tan x} = \int (\sec x - \tan x) dx \]

We can integrate each term separately:

\[ \int (\sec x - \tan x) dx = \int \sec x \, dx - \int \tan x \, dx \]

We need to know the standard integrals for \(\sec x\) and \(\tan x\):

• The integral of \(\sec x\) is \(\int \sec x \, dx = \ln |\sec x + \tan x| + C_1\).
• The integral of \(\tan x\) is \(\int \tan x \, dx = \ln |\sec x| + C_2\) or \(\int \tan x \, dx = -\ln |\cos x| + C_2\). Note that \(-\ln |\cos x| = \ln |(\cos x)^{-1}| = \ln |\sec x|\), so both forms are equivalent. We will use \(\ln |\sec x|\).

Substituting these standard integrals into our expression:

\[ \int \sec x \, dx - \int \tan x \, dx = (\ln |\sec x + \tan x| + C_1) - (\ln |\sec x| + C_2) \]

Combining the constants of integration \(C_1\) and \(C_2\) into a single constant \(C = C_1 - C_2\), we get:

\[ \int \frac{dx}{\sec x+\tan x} = \ln |\sec x + \tan x| - \ln |\sec x| + C \]

Comparing with Options

Let's compare our result with the given options:

• Option 1: \(\ln (\sec x) + \ln |\sec x + \tan x| + c\) - Does not match the signs.
• Option 2: \(\ln (\sec x) - \ln |\sec x + \tan x| + c\) - Does not match the signs.
• Option 3: \(\sec x \tan x - \ln |\sec x - \tan x| + c\) - Different form entirely.
• Option 4: \(\ln |\sec x + \tan x| - \ln |\sec x| + c\) - This matches our derived solution.

Thus, the integral is equal to \(\ln |\sec x + \tan x| - \ln |\sec x| + C\).

Revision Table: Key Integrals Used

Additional Information: Alternative Approach and Properties

While multiplying by the conjugate is efficient here, another way to approach integrals with \(\sec x\) and \(\tan x\) is to convert everything into sines and cosines:

\[ \frac{1}{\sec x + \tan x} = \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1}{\frac{1+\sin x}{\cos x}} = \frac{\cos x}{1+\sin x} \]

Now, the integral is \(\int \frac{\cos x}{1+\sin x} dx\). This can be solved using a simple substitution. Let \(u = 1+\sin x\). Then \(du = \cos x \, dx\).

The integral becomes \(\int \frac{du}{u}\), which is \(\ln |u| + C\).

Substituting back \(u = 1+\sin x\), we get \(\ln |1+\sin x| + C\).

Is this result equivalent to \(\ln |\sec x + \tan x| - \ln |\sec x| + C\)?

Using logarithm properties, \(\ln |\sec x + \tan x| - \ln |\sec x| = \ln \left| \frac{\sec x + \tan x}{\sec x} \right|\).

Let's simplify the argument of the logarithm:

\[ \frac{\sec x + \tan x}{\sec x} = \frac{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x}} = \frac{\frac{1+\sin x}{\cos x}}{\frac{1}{\cos x}} = \frac{1+\sin x}{\cos x} \]

So, \(\ln |\sec x + \tan x| - \ln |\sec x| = \ln \left| \frac{1+\sin x}{\cos x} \right|\).

This doesn't immediately look like \(\ln |1+\sin x|\). However, consider \(\left| \frac{1+\sin x}{\cos x} \right|\). If \(1+\sin x > 0\), this is \(\left| \frac{(1+\sin x)(1-\sin x)}{\cos x (1-\sin x)} \right| = \left| \frac{1-\sin^2 x}{\cos x (1-\sin x)} \right| = \left| \frac{\cos^2 x}{\cos x (1-\sin x)} \right| = \left| \frac{\cos x}{1-\sin x} \right|\).

Let's re-check the first form: \(\ln |1+\sin x|\). This form is simpler and likely preferred if allowed by the context (e.g., domain of integration). Both forms represent the same family of functions (differing potentially by a constant or defined on different intervals depending on where \(\sec x\) and \(\tan x\) are defined). The provided options lead us to the form involving \(\sec x\) and \(\tan x\).

The identity used \(\sec^2 x - \tan^2 x = 1\) is a direct consequence of dividing the identity \(\sin^2 x + \cos^2 x = 1\) by \(\cos^2 x\).

Logarithm property used: \(\ln a - \ln b = \ln \frac{a}{b}\).