Video Transcript
Is the integral from zero to 𝜋 of the tan of 𝑥 with respect to 𝑥 an improper integral?
In this question, we’re given a definite integral and we’re asked to determine whether this is an improper integral. So, the first thing we’re going to need to do is remember what we mean by an improper integral. We call a definite integral an improper integral if either of the following two things are true. First, if either of the two limits of integration are positive or negative ∞, then we call our integral improper. And this is because if either the upper or lower limit of integration are positive or negative ∞, we’ll need to calculate this integral by using limits.
Second, if the integrand is not continuous on the entire interval of integration, then we call our integral improper. And once again, this is because to evaluate integrals where our integrand is not continuous on the entire interval of integration, we’re going to need to use limits. In the second case, we would need to split our integral at the points of discontinuity.
So, these are the two ways that our integral could be improper. Let’s see if either of these hold to the integral given to us in the question. First, we need to check if either of the limits of integration are equal to positive or negative ∞. In the integral given to us in the question, the upper limit of integration is 𝜋 and the lower limit of integration is zero, and neither of these are positive or negative ∞. So, the first condition is not true for this Integral.
What about the second condition? We need to check if our integrand is continuous across the entire interval of integration. First, let’s recall what we mean by the interval of integration. In our case, the definite integral from zero to 𝜋 of the tan of 𝑥 with respect to 𝑥 means the area under the curve the tan of 𝑥 between the lines 𝑥 is equal to zero and 𝑥 is equal to 𝜋. And this shape will have a height of the tan of 𝑥, and its values of 𝑥 will range from zero to 𝜋. We call the closed interval from zero to 𝜋 the interval of integration.
So, to determine if this integral is improper or not, we need to determine if the tan of 𝑥 is continuous on the closed interval from zero to 𝜋. And in fact, there’s a lot of different ways of approaching this. We’ll use the trigonometric identity the tan of 𝑥 is equivalent to the sin of 𝑥 divided by the cos of 𝑥. By writing the tan of 𝑥 in this form, we can see it’s the quotient of two continuous functions. We know the sin of 𝑥 and the cos of 𝑥 are continuous for all real values of 𝑥, and the quotient of continuous functions is continuous across its entire domain.
So, all we need to do is ask the question “when is the sin of 𝑥 divided by the cos of 𝑥 not defined?” It will be defined everywhere unless we’re dividing by zero. So, we would want to ask the question “when is the cos of 𝑥 equal to zero?” We could find all the solutions to this equation; however, remember, we’re only interested in is the tan of 𝑥 continuous on the closed interval from zero to 𝜋.
So, in fact, all we need to do is take the inverse cos of both sides of this equation. And of course, the inverse cos of zero is equal to 𝜋 by two. So, the cos of 𝜋 by two is equal to zero. And therefore, the tan of 𝜋 by two is undefined. And of course, for a function to be continuous at a point, it must be defined at this point. In other words, because the tan of 𝜋 by two is not defined, the tan of 𝑥 is not continuous at 𝑥 is equal to 𝜋 by two. And of course, 𝜋 by two is on the closed interval from zero to 𝜋. This means the tan of 𝑥 is not continuous on the closed interval from zero to 𝜋. And in turn, the second condition for improper integrals does hold in this case.
Therefore, we’ve shown to answer the question is the integral from zero to 𝜋 of the tan of 𝑥 with respect to 𝑥 an improper integral, the answer is yes, because it is not continuous on the entire interval of integration.
