Definite and Improper Integral Calculator

Your input: calculate $$$\int_{\frac{\pi}{6}}^{\frac{e \pi}{3}}\left( \frac{1}{f \cos{\left(x \right)}} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{\frac{1}{f \cos{\left(x \right)}} d x}=\frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{f}$$$ (for steps, see indefinite integral calculator)

The interval of integration contains the point $$$\frac{\pi}{2}$$$, which is not in the domain of the integrand, so this is an improper integral of type 2.

Therefore, divide the interval into the following subintervals: $$$\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$$$, $$$\left(\frac{\pi}{2}, \frac{e \pi}{3}\right)$$$. Evaluate the integral over each subinterval.

To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (is not in the domain of the function).

$$$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\left( \frac{1}{f \cos{\left(x \right)}} \right)dx=\lim_{x \to \frac{\pi}{2}}\left(\frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{f}\right)-\left(\frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{f}\right)|_{\left(x=\frac{\pi}{6}\right)}=\infty - \frac{\ln{\left(\sqrt{3} \right)}}{f}$$$

Since the value of the integral is not finite, the value of the initial integral is not finite either. Thus, the integral is divergent.

Answer: the integral is divergent.