Question Video: Evaluating the Improper Integral of a Discontinuous Function by Using Integration by Substitution Mathematics

Video Transcript

The integral between zero and 𝜋 over two of cos 𝜃 divided by the square root of sin 𝜃 with respect to 𝜃 is convergent. What does it converge to?

Straight away, we can notice that the question has told us our integral is convergent. This gives us a hint that we’re looking at an improper integral. Now, since we don’t see any infinities as our limits of integration, it’s also implied that our integrand has a discontinuity on the closed interval between zero and 𝜋 over two, which are the limits of the integral. Let’s look for this discontinuity. Since the integrand that we’ve been given is a quotient, we should try and find where its denominator is equal to zero. The square root of sin 𝜃 is zero, where sin 𝜃 itself is also equal to zero. This occurs when 𝜃 is equal to zero plus 𝑛𝜋, which is of course just 𝑛𝜋. And 𝑛 here is an integer. So here we see that our denominator is equal to zero when 𝜃 is equal to any of the following. The only one of these values which occurs over the interval of our integration is zero itself.

Okay, for due diligence, let’s now check what happens to our numerator when 𝜃 is equal to zero. Our numerator is cos 𝜃. So this becomes cos of zero, which is equal to one. We do this to check that when 𝜃 equals zero, our function is not equal to the indeterminate form of zero over zero. Since we would then be looking at a removable discontinuity. In actual fact, when 𝜃 is equal to zero, our function is equal to one over zero, which is the case of an infinite discontinuity. Okay. We’ve now confirmed that we have an improper integral. And we have an infinite discontinuity at the lower limit of integration. The definition of an improper integral specifically for this case tells us that if 𝑓 is a continuous function on the interval which is open at 𝑎 and closed at 𝑏 and discontinuous at 𝑎. Then the integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑡 approaches 𝑎 from the right of the integral between 𝑡 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥. If this limit exists and is finite.

Let’s now apply this to our question. The integral that we’ve been given is equal to the limit as 𝑡 approaches zero from the right of the same integral, but now between the limits of 𝑡 and 𝜋 over two. Now, at this point, we might notice that our integral is not entirely trivial. And we’ll need to put in some work to evaluate it. To move forward, we’re gonna be using a 𝑢-substitution, specifically 𝑢 is equal to sin 𝜃. From this, we differentiate to find that d𝑢 by d𝜃 is equal to cos 𝜃. An equivalent statement to this is that d𝑢 is equal to cos 𝜃 d𝜃. We now do the following with our substitution. We replace sin 𝜃 with 𝑢. And we replace cos 𝜃 d𝜃 with d𝑢. We’re then left with the integral of one over the square root of 𝑢 with respect to 𝑢. Another way of expressing this integrand is 𝑢 to the power of negative a half. Evaluating this, we get two 𝑢 to the power of a half, which, is of course, two times the square root of 𝑢. We can now go back to our original substitution and replace our 𝑢 with sin 𝜃. We therefore have that our integral is equal to two times the square root of sin 𝜃, of course, plus the constant of integration.

Now that we’ve done the legwork on our integral, let’s use this result to move forward with our question. We use the antiderivative that we’ve just found for our integral. And we get the following. We now input our limits of integration 𝑡 and 𝜋 over two. And we’re left with the following limit. We can now take a direct substitution approach to our limit. And we can evaluate since sin of 𝜋 over two is equal to one. And sin of zero is equal to zero. We’re then left with two times the square of one minus two times the square root of zero. Of course, the square root of one is just one. And this entire term is just zero. The answer that we’re left with is therefore two. Upon reaching this result, we have answered our question. We used our definition to find that the integral given in the question converges to a value of two.