1. **State the problem:** We are given the curve defined by the integral $$x = \int_0^y \sqrt{\sec^2 t - 1} \, dt$$ for $$-\frac{\pi}{3} \leq y \leq \frac{\pi}{4}$$ and asked to find the length of this curve. 2. **Simplify the integrand:** Recall the trigonometric identity $$\sec^2 t - 1 = \tan^2 t$$. Therefore, $$\sqrt{\sec^2 t - 1} = \sqrt{\tan^2 t} = |\tan t|$$. 3. **Rewrite the curve:** The curve is given by $$x = \int_0^y |\tan t| \, dt$$. 4. **Find the derivative:** Since $$x$$ is defined as an integral with respect to $$y$$, by the Fundamental Theorem of Calculus, $$\frac{dx}{dy} = |\tan y|$$. 5. **Parametric form:** The curve can be represented parametrically as $$x(y) = \int_0^y |\tan t| \, dt, \quad y = y$$. 6. **Length of the curve formula:** The length $$L$$ of a curve defined parametrically by $$x = x(y)$$ and $$y = y$$ over $$y \in [a,b]$$ is $$L = \int_a^b \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy$$. 7. **Substitute $$\frac{dx}{dy}$$:** $$L = \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}} \sqrt{|\tan y|^2 + 1} \, dy = \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}} \sqrt{\tan^2 y + 1} \, dy$$. 8. **Simplify the integrand:** Using the identity $$1 + \tan^2 y = \sec^2 y$$, $$L = \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}} |\sec y| \, dy$$. 9. **Evaluate the integral:** On the interval $$[-\frac{\pi}{3}, \frac{\pi}{4}]$$, $$\cos y$$ is positive for $$y \in (-\frac{\pi}{3}, \frac{\pi}{2})$$, so $$\sec y = \frac{1}{\cos y} > 0$$. Therefore, $$L = \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}} \sec y \, dy$$. 10. **Integral of $$\sec y$$:** $$\int \sec y \, dy = \ln |\sec y + \tan y| + C$$. 11. **Calculate definite integral:** $$L = \left[ \ln |\sec y + \tan y| \right]_{-\frac{\pi}{3}}^{\frac{\pi}{4}} = \ln |\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| - \ln |\sec (-\frac{\pi}{3}) + \tan (-\frac{\pi}{3})|$$. 12. **Evaluate values:** - $$\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$$ - $$\tan \frac{\pi}{4} = 1$$ - $$\sec (-\frac{\pi}{3}) = \sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2$$ - $$\tan (-\frac{\pi}{3}) = -\tan \frac{\pi}{3} = -\sqrt{3}$$ 13. **Substitute:** $$L = \ln |\sqrt{2} + 1| - \ln |2 - \sqrt{3}| = \ln \frac{\sqrt{2} + 1}{2 - \sqrt{3}}$$. 14. **Rationalize denominator:** $$2 - \sqrt{3} = (2 - \sqrt{3}) \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(2)^2 - (\sqrt{3})^2}{2 + \sqrt{3}} = \frac{4 - 3}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}}$$ Therefore, $$\frac{\sqrt{2} + 1}{2 - \sqrt{3}} = (\sqrt{2} + 1)(2 + \sqrt{3})$$. 15. **Multiply out:** $$(\sqrt{2} + 1)(2 + \sqrt{3}) = 2\sqrt{2} + \sqrt{6} + 2 + \sqrt{3}$$. 16. **Final answer:** $$L = \ln(2\sqrt{2} + \sqrt{6} + 2 + \sqrt{3})$$. **Summary:** The length of the curve is $$\boxed{L = \ln(2\sqrt{2} + \sqrt{6} + 2 + \sqrt{3})}$$.