Questions: =∫ from 2 to 3 sec θ / tan^2 θ

Transcript text: $=\int_{2}^{3} \frac{\sec \theta}{\tan ^{2} \theta}$

Solution

Solution Steps

To solve the integral $\int_{2}^{3} \frac{\sec \theta}{\tan ^{2} \theta} \, d\theta$, we can use a trigonometric identity to simplify the integrand. We know that $\tan^2 \theta = \sec^2 \theta - 1$, so we can rewrite the integrand in terms of $\sec \theta$ and simplify. After simplification, we can integrate the resulting expression over the given limits.

Step 1: Set Up the Integral

We start with the integral

\[ I = \int_{2}^{3} \frac{\sec \theta}{\tan^2 \theta} \, d\theta. \]

Step 2: Simplify the Integrand

Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, we can express the integrand as

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

However, we can directly integrate the expression as it is.

Step 3: Evaluate the Integral

The result of the integral is

\[ I = -\frac{1}{\sin(3)} + \frac{1}{\sin(2)}. \]