Question

Find: \[ \int \sec^3 \theta \, d\theta \]

Hint:

When integrating \( \sec^3 \theta \), it's helpful to use trigonometric identities and substitution to simplify the expression. Pay attention to how derivatives of trigonometric functions can be leveraged in substitution.

Answer 1

We are asked to evaluate the integral: \[ I = \int \sec^3 \theta \, d\theta \] Step 1: Use the identity for \( \sec^3 \theta \) 
We can express \( \sec^3 \theta \) as: \[ \sec^3 \theta = \sec^2 \theta \cdot \sec \theta \] Thus, the integral becomes: \[ I = \int \sec^2 \theta \cdot \sec \theta \, d\theta \] Step 2: Substitute and simplify the integral 
Now, let's use the substitution method: Let \( u = \sec \theta \), then \( \frac{du}{d\theta} = \sec \theta \tan \theta \), so we can rewrite the integral: \[ I = \sec \theta \int \sec^2 \theta \, d\theta - \int \frac{d(\sec \theta)}{d\theta} \left( \int \sec^2 \theta \, d\theta \right) d\theta \] Step 3: Finish the integration
After simplifying and solving: \[ I = \sec \theta \tan \theta - \int \sec^3 \theta \, d\theta + \int \sec \theta \, d\theta \] \[ I = \frac{1}{2} \left( \sec \theta \tan \theta + \log |\sec \theta + \tan \theta| + c \right) \] Thus, the solution to the integral is: \[ I = \frac{1}{2} \left( \sec \theta \tan \theta + \log |\sec \theta + \tan \theta| + c \right) \] Step 4: Correct Answer:
The correct solution to the integral is: \[ I = \frac{1}{2} \left( \sec \theta \tan \theta + \log |\sec \theta + \tan \theta| + c \right) \]