91Ó°ÊÓ

We know that: $$\sec^2\theta - 1 = \tan^2\theta$$ So, the given integral can be rewritten as: $$\int_{-\pi/3}^{\pi/3} \sqrt{\tan^2\theta} d\theta$$ Since \(\tan \theta\) is an odd function, we have: $$\int_{-\pi/3}^{\pi/3} \sqrt{\tan^2\theta} d\theta = \int_{-\pi/3}^{\pi/3} |\tan\theta| d\theta$$

Let's use the substitution method with: $$u = \tan\theta \Rightarrow du = \sec^2\theta d\theta$$ In terms of \(u\), the limits of integration change as well. When \(\theta=-\pi/3\), \(u=-\sqrt{3}\), and when \(\theta=\pi/3\), \(u=\sqrt{3}\). Hence, our integral becomes: $$\int_{-\sqrt{3}}^{\sqrt{3}} |u| du$$ Now we can split the integral into two parts: $$\int_{-\sqrt{3}}^{\sqrt{3}} |u| du = \int_{-\sqrt{3}}^{0}-u du + \int_{0}^{\sqrt{3}}u du$$ Calculating these integrals, we have: $$\begin{aligned} \int_{-\sqrt{3}}^{0}-u du + \int_{0}^{\sqrt{3}}u du &= -\frac{1}{2} u^2\Big|_{-\sqrt{3}}^{0} + \frac{1}{2}u^2\Big|_{0}^{\sqrt{3}} \\ &= -\frac{1}{2}(0^2-(-\sqrt{3})^2) + \frac{1}{2}(\sqrt{3}^2 - 0^2)\\ &= -\frac{1}{2}(0 - 3) + \frac{1}{2}(3 - 0)\\ &= \frac{3}{2} \end{aligned} $$ So the final answer is: $$\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} \theta-1} d \theta = \frac{3}{2}$$