Let's solve the definite integral: \[ \int_{}^{\frac{\pi}{6}} \sec 2\theta \tan 2\theta \, d\theta \] --- ## Step 1: Find the Indefinite Integral Recall the derivative: \[ \frac{d}{dx}(\sec x) = \sec x \tan x \] Now, apply substitution: Let \( u = 2\theta \), hence \( du = 2 d\theta \) or \( d\theta = \frac{du}{2} \). But since the integrand already matches the derivative form, let's directly integrate: \[ \int \sec 2\theta \tan 2\theta \, d\theta = \frac{1}{2} \sec 2\theta + C \] **Explanation:** We use the chain rule; the derivative of \(\sec 2\theta\) with respect to \(\theta\) is \(2 \sec 2\theta \tan 2\theta\), so our integral gives half of \(\sec 2\theta\). --- ## Step 2: Evaluate the Definite Integral Plug in the limits: \[ \left. \frac{1}{2} \sec 2\theta \right|_{}^{\frac{\pi}{6}} = \frac{1}{2} \left[\sec\left(2 \cdot \frac{\pi}{6}\right) - \sec()\right] \] \[ = \frac{1}{2} \left[\sec\left(\frac{\pi}{3}\right) - \sec()\right] \] \[ \sec\left(\frac{\pi}{3}\right) = 2, \quad \sec() = 1 \] \[ = \frac{1}{2} (2 - 1) = \frac{1}{2} (1) = \frac{1}{2} \] **Explanation:** We substitute the upper and lower limits into our result for the antiderivative and subtract. --- ## Final Answer \[ \boxed{\frac{1}{2}} \]