Problem 36 - Essential Calculus

Question

Find or evaluate the integral. $$ \int \frac{1}{\sec \theta-\tan \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( ln |\sec \theta - \tan \theta | + C \)

Step 1: Trigonometric substitution

First, substitute \( \sec \theta - \tan \theta \) with \( u \). This gives: \( \sec \theta - \tan \theta = u \). Let's differentiate the both sides with respect to \( \theta \) to find the relation in terms of \( d\theta \) and \( du \). Differentiating gives: \( \sec \theta \tan \theta - \sec^2 \theta = du/d\theta \).

Step 2: Express \( d\theta \) in terms of \( du \)

Knowing that \( d\theta = du / ( \sec \theta \tan \theta - \sec^2 \theta) \), we'll substitute \( d\theta \) into our integral in terms of \( u \) to get: \( \int 1/u \times du / ( \sec \theta \tan \theta - \sec^2 \theta) \)

Step 3: Simplify the integral

The integral simplifies to \( \int du/u = ln|u| + C \), applying the rules of integration.

Step 4: Substitute the original variable back

Finally, substitute \( u = \sec \theta - \tan \theta \) into the result to get the final answer as: \( ln |\sec \theta - \tan \theta | + C \)