Question

Evaluate the definite integral of the transcendental function. Use a graphing utility to verify your result. $$ \int_{-\pi / 3}^{\pi / 3} 4 \sec \theta \tan \theta d \theta $$

Short answer

The result of the definite integral \( \int_{-\pi / 3}^{\pi / 3} 4 \sec \theta \tan \theta d \theta \) is 48

Step-by-step solution

Identify the Integral

The given integral is \( \int_{-\pi / 3}^{\pi / 3} 4 \sec \theta \tan \theta d \theta \) which is a product of secant and tangent functions.

Use trigonometric identities and derivatives

The derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \). So, we could apply substitution method. Setting \( u = \sec \theta \), we get that \( du = \sec \theta \tan \theta d \theta \).

Substitute and solve

After the substitution, the initial integral converts to \( \int_{-\sqrt{3}^2}^{\sqrt{3}^2}4u du \) which simplifies to \( [2u^2]_{-\sqrt{3}^2}^{\sqrt{3}^2} \).

Evaluate the definite integral

By substituting upper and lower limits, we get \( [2(\sqrt{3})^{2}^2] - [2(-\sqrt{3})^{2}^2] \) which simplifies to 24 - (-24), equal to 48.

Verification using Graphing utility

Verify this result by plotting Y= \( 4 \sec \theta \tan \theta \) in the interval -π/3 to π/3. Compute the definite integral for this range. The calculated area under the curve is 48, confirming our result.