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

=∫ from 2 to 3 sec θ / tan^2 θ
Transcript text: $=\int_{2}^{3} \frac{\sec \theta}{\tan ^{2} \theta}$
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Solution

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Solution Steps

To solve the integral 23secθtan2θdθ\int_{2}^{3} \frac{\sec \theta}{\tan ^{2} \theta} \, d\theta, we can use a trigonometric identity to simplify the integrand. We know that tan2θ=sec2θ1\tan^2 \theta = \sec^2 \theta - 1, so we can rewrite the integrand in terms of secθ\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=23secθtan2θdθ. I = \int_{2}^{3} \frac{\sec \theta}{\tan^2 \theta} \, d\theta.

Step 2: Simplify the Integrand

Using the identity tan2θ=sec2θ1\tan^2 \theta = \sec^2 \theta - 1, we can express the integrand as

secθtan2θ=secθsec2θ1. \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=1sin(3)+1sin(2). I = -\frac{1}{\sin(3)} + \frac{1}{\sin(2)}.

Final Answer

Thus, the final result of the integral is

I=1sin(3)+1sin(2) \boxed{I = -\frac{1}{\sin(3)} + \frac{1}{\sin(2)}}

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