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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!