Problem 39 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d x}{\sec x-1}$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the following function: $$\int \frac{dx}{\sec{x}-1}$$ Answer: $$\int \frac{dx}{\sec{x}-1} = \ln{|\sin{x}|} + C$$

Step 1: Rewrite the integrand using trigonometric identities

Recall the trigonometric identity for sec(x): $$\sec{x} = \frac{1}{\cos{x}}$$ Substitute this into the integrand: $$\int \frac{d x}{\sec x-1} = \int \frac{d x}{\frac{1}{\cos x} - 1}$$

Step 2: Simplify the integrand

To simplify the integrand, multiply both the numerator and the denominator by the conjugate of the denominator, which is $$\cos x + 1$$: $$\int \frac{d x}{\frac{1}{\cos x} - 1} = \int \frac{(\cos x + 1) d x}{\left(\frac{1}{\cos x} - 1\right) (\cos x + 1)}$$ Multiply, and we get: $$\int \frac{(\cos x + 1) d x}{1 - \cos^2 x}$$

Step 3: Observe the Pythagorean relation

Notice that we have in the denominator the Pythagorean identity $$1 - \cos^2 x = \sin^2 x$$. So, we can rewrite the integral as: $$\int \frac{(\cos x + 1) d x}{\sin^2 x}$$

Step 4: Use substitution method

Let's make a simple substitution: $$u = \sin{x} \Rightarrow d u = \cos{x} d x$$ Now we have: $$\int \frac{(\cos x + 1) d x}{\sin^2 x} = \int \frac{u d u}{u^2}$$

Step 5: Integrate with respect to u

Now the integration becomes straightforward. Simplify the integral and integrate: $$\int \frac{u d u}{u^2} = \int \frac{d u}{u}$$ Using the elementary integration rule, we have: $$\int \frac{d u}{u} = \ln{|u|} + C = \ln{|\sin{x}|} + C$$ So, the final answer is: $$\int \frac{d x}{\sec x-1} = \ln{|\sin{x}|} + C$$