Paul's Online Notes
Home / Calculus I / Integrals / Computing Definite Integrals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5-7 : Computing Definite Integrals

11. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

$\int_{0}^{\pi }{{\sec \left( z \right)\tan \left( z \right) - 1\,dz}}$ Show Solution

Be careful with this integral. Recall that,

$\sec \left( z \right) = \frac{1}{{\cos \left( z \right)}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\tan \left( z \right) = \frac{{\sin \left( z \right)}}{{\cos \left( z \right)}}$

Also recall that $$\cos \left( {\frac{\pi }{2}} \right) = 0$$ and that $$x = \frac{\pi }{2}$$ is in the interval we are integrating over, $$\left[ {0,\pi } \right]$$ and hence is not continuous on this interval.

Therefore, this integral cannot be done.

It is often easy to overlook these kinds of division by zero problems in integrands when the integrand is not explicitly written as a rational expression. So, be careful and don’t forget that division by zero can sometimes be “hidden” in the integrand!