# Video: Determining Whether the Integrals of Trigonometric Functions with Exponents Are Convergent or Divergent

Determine whether the integral ∫_(0) ^(𝜋/2) tan² 𝜃 d𝜃 is convergent or divergent.

04:28

### Video Transcript

Determine whether the integral from zero to 𝜋 by two of the tan squared of 𝜃 with respect to 𝜃 is convergent or divergent.

The question gives us a definite integral. And it wants us to determine if this definite integral is convergent or divergent. The first thing we notice about our definite integral is neither the upper limit or lower limit contain ∞. This means the only way our integral can be improper is if our integrand is not continuous on this interval. So, let’s check the continuity of our integrand on the closed interval from zero to 𝜋 by two.

First, the tangent function is a trigonometric function. And we know all trigonometric functions are continuous on their entire domain. On the closed interval from zero to 𝜋 by two, the only place the tangent function is not defined is when 𝜃 is equal to 𝜋 by two. So, tan 𝜃 is continuous on our entire interval except when 𝜃 is 𝜋 by two.

Next, we know the square function is a polynomial. This means it’s continuous for all real values of 𝑥. This tells us that the tan of 𝜃 all squared is the composition of two continuous functions. This means it’s continuous on its entire domain. And the domain of the tan squared of 𝜃 is the domain of the tan of 𝜃. So, the tan squared of 𝜃 is continuous on our entire closed interval except when 𝜃 is 𝜋 by two. So, our integrand is continuous on our integral everywhere except at the upper limits of our integral.

This tells us we’re going to need to use one of our rules for improper integrals. We recall if 𝑓 is a continuous function on the open interval from 𝑎 to 𝑏, 𝑓 is continuous at 𝑎, and 𝑓 has a discontinuity at 𝑏. Then we can evaluate the definite integral from 𝑎 to 𝑏 of 𝑓 of 𝜃 with respect to 𝜃 as the limit as 𝑡 approaches 𝑏 from the left of the integral from 𝑎 to 𝑡 of 𝑓 of 𝜃 with respect to 𝜃. And this, of course, is only true if this limit exists. Then, we call our integral convergent. If this limit does not exist, we call our integral divergent.

We’ve already shown this is what we have in the integral given to us in the question. We’ll set 𝑎 to be zero, 𝑏 to be 𝜋 by two, and 𝑓 of 𝜃 to be our integrand the tan squared of 𝜃. So, by using our rule for improper integrals, we have the integral from zero to 𝜋 by two of the tan squared of 𝜃 with respect to 𝜃. Is equal to the limit as 𝑡 approaches 𝜋 by two from the left of the integral from zero to 𝑡 of the tan squared of 𝜃 with respect to 𝜃. And, of course, this is only true if this limit exists.

To help us evaluate this integral, we recall that the tan squared of 𝜃 is equivalent to the sec squared of 𝜃 minus one. We get this from the Pythagorean identity. Using this to rewrite our integrand, we now need to find the limit as 𝑡 approaches 𝜋 by two from the left of the integral from zero to 𝑡 of the sec squared of 𝜃 minus one with respect to 𝜃.

But, now, we can see that our integral is in a form which we can evaluate. We know the integral of the sec squared of 𝜃 with respect to 𝜃 is equal to the tan of 𝜃 plus the constant of integration 𝐶. And, of course, we can evaluate the integral of negative one. It’s equal to negative 𝜃 plus the constant of integration. This gives us the limit as 𝑡 approaches 𝜋 by two from the left of the tan of 𝜃 minus 𝜃 evaluated at the limits of our integral, zero and 𝑡.

We’ll now evaluate this expression at the limits of our integral. We get the limit as 𝑡 approaches 𝜋 by two from the left of the tan of 𝑡 minus 𝑡 minus the tan of zero plus zero. We can then simplify this expression. We know the tan of zero is equal to zero. And adding zero won’t change the value of our limit. So, this gives us the limit as 𝑡 approaches 𝜋 by two from the left of the tan of 𝑡 minus 𝑡.

And, now, we see we have a problem. As 𝑡 is approaching 𝜋 by two from the left, we know that negative 𝑡 is getting closer and closer to negative 𝜋 by two. However, the tangent of 𝑡 has an infinite discontinuity at 𝑡 is equal to 𝜋 by two. In this case, we know as 𝑡 approaches 𝜋 by two from the left, the tan of 𝑡 is going to get larger and larger. In fact, it’s going to grow without bound. This means our entire limit is growing without bound. So, this limit is equal to ∞.

And remember, this means that our limit does not converge. And this limit not converging is the same as saying that our integral does not converge. This means that our integral is divergent.

Therefore, we’ve shown the integral from zero to 𝜋 by two of the tan squared of 𝜃 with respect to 𝜃 is divergent.