Video: Finding the Definite Integral of an Inverse Trigonometric Function

Calculate ∫_(0)^(1) tan⁻¹ (π‘₯) dπ‘₯.

07:21

Video Transcript

Calculate the definite integral from zero to one of the inverse tan of π‘₯ with respect to π‘₯.

We’re given a definite integral we need to calculate. Our integrand is the inverse tan of π‘₯ with respect to π‘₯. The first thing we always need to check when we’re calculating an indefinite integral is to check that our integrand is continuous over the limits of integration. In this case, we know the inverse tan of π‘₯ is continuous for all real values of π‘₯. So we don’t need to worry about this.

But we do have a problem. We don’t know how to integrate the inverse tan of π‘₯. We only know how to differentiate this with respect to π‘₯. We do know a few different methods to try and integrate functions. For example, we might be tempted to try using integration by substitution. The only substitution that seems it would make sense would be to set π‘₯ equal to the tan of πœƒ. But if we did this, then we would see dπ‘₯ by dπœƒ is the derivative of the tan of πœƒ with respect to πœƒ. And we know this is the sec squared of πœƒ.

And if we were to then use our substitution just in the definite integral, we would get the inverse tan of the tan of πœƒ times the sec squared of πœƒ with respect to πœƒ. And of course, the inverse tan of the tan of πœƒ is just πœƒ, so we would need to evaluate the integral of πœƒ times the sec squared of πœƒ with respect to πœƒ. And this is a very complicated-looking integral. We might be able to work this out. However, it would be very easy to make a mistake. Instead, we’re going to try one of our other rules. We’ll try using integration by parts.

First, we recall integration by parts tells us the integral of 𝑒 multiplied by 𝑣 prime with respect to π‘₯ is equal to 𝑒 times 𝑣 minus the integral of 𝑣 times 𝑒 prime with respect to π‘₯. So this gives us a method of integrating the product of 𝑒 and 𝑣 prime with respect to π‘₯. However, we can immediately see two problems. First, our integrand is not the product of two functions. It’s just the inverse tan of π‘₯. Secondly, this gives us an indefinite integral. We want to calculate a definite integral. So it might seem like we can’t use integration by parts to solve this question. However, we can actually overcome both of these obstacles by using what we know about integration.

Let’s start with the first problem that our integrand is not the product of two functions. Well, we can fix this by just writing this integral as the definite integral from zero to one of the inverse tan of π‘₯ multiplied by one with respect to π‘₯. Now, our integrand is the product of two functions.

But we still have the second problem. We’re still calculating a definite integral. In fact, we can get around this problem by using what we know about definite integrals. First, we’ll use integration by parts only on the indefinite integral. Then if our integration by parts method works, we would get some function, capital 𝐹 of π‘₯, plus a constant of integration 𝐢. The important thing here is capital 𝐹 of π‘₯ would be an antiderivative of the inverse tan of π‘₯. But then, if we have an antiderivative of our function, we can calculate its definite integral. The definite integral of the inverse tan of π‘₯ with respect to π‘₯ would be capital 𝐹 of π‘₯ evaluated at the limits of integration, zero and one.

We’re now almost ready to try and evaluate this integral by using integration by parts. The last thing we need to do is choose our functions 𝑒 and 𝑣 prime. There’s a few different ways of doing this. For example, we could use the LIATE method. Since there are no logarithmic functions, this tells us we should set our function 𝑒 to be the inverse trigonometric function, the inverse tan of π‘₯. However, in this case, this isn’t necessary. We’ve already said we don’t know how to integrate the inverse tan of π‘₯. This means we have to pick our function 𝑣 prime as one.

So we’ve set 𝑒 to be the inverse tan of π‘₯ and 𝑣 prime equal to one. We now need to find expressions for 𝑒 prime and 𝑣. Let’s start with 𝑒 prime. That’s the derivative of the inverse tan of π‘₯ with respect to π‘₯. This is a standard inverse trigonometric derivative result we should commit to memory. It’s equal to one divided by one plus π‘₯ squared. Next, to find 𝑣, we just need an antiderivative of one. We know this is just equal to π‘₯. We can now use our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime to evaluate our integration by parts.

We get the inverse tan of π‘₯ multiplied by π‘₯ minus the integral of π‘₯ times one over one plus π‘₯ squared with respect to π‘₯. The first thing we’ll do is simplify our integrand to give us π‘₯ divided by one plus π‘₯ squared. So to find an antiderivative we can use, we now need to integrate π‘₯ divided by one plus π‘₯ squared with respect to π‘₯. The easiest way to evaluate this integral is to notice a pattern. If we set the function 𝐹 of π‘₯ to be the function in our denominator, that’s one plus π‘₯ squared.

Then if we calculate 𝐹 prime of π‘₯, we get two π‘₯. We see this is a scalar multiple of our numerator. And this is in a form we should recognize. We know the integral of 𝐹 prime of π‘₯ divided by 𝐹 of π‘₯ with respect to π‘₯ is equal to the natural logarithm of the absolute value of 𝐹 of π‘₯ plus a constant of integration 𝐢. In this case, we don’t quite have 𝐹 prime of π‘₯ in our numerator. So we’ll multiply our integrand by two and then multiply the entire integral by one-half.

Now, we have the integral of 𝐹 prime of π‘₯ divided by 𝐹 of π‘₯ with respect to π‘₯. So we can use this formula. So by using this, we now have π‘₯ times the inverse tan of π‘₯ minus one-half multiplied by the natural logarithm of the absolute value of one plus π‘₯ squared plus a constant of integration 𝐢. But remember, we’re using this to find an antiderivative which we will eventually use in a definite integral. So, in fact, we don’t need our constant of integration.

So we can remove this. And now, we found an antiderivative of the inverse tan of π‘₯. Now that we found our antiderivative, we’re finally ready to evaluate the integral from zero to one of the inverse tan of π‘₯ with respect to π‘₯. It’s equal to π‘₯ times the inverse tan of π‘₯ minus one-half times the natural logarithm of the absolute value of one plus π‘₯ squared evaluated at limits of integration π‘₯ is equal to zero and π‘₯ is equal to one.

Evaluating this at the limits of integration zero and one, we get one times the inverse tan of one minus one-half multiplied by the natural logarithm of the absolute value of one plus one squared minus zero times the inverse tan of zero minus a half multiplied by the natural logarithm of the absolute value of one plus zero squared. And now, we can just start evaluating.

First, our first term, the inverse tan of one, is πœ‹ by four. And πœ‹ by four multiplied by one is just equal to πœ‹ by four. So our first term is just πœ‹ by four. In our second term, one plus one squared is equal to two. And the absolute value of two is just equal to two. So our second term simplifies to give us negative the natural logarithm of two all divided by two. Our third term has a factor of zero. And our fourth term has a factor of the natural logarithm of the absolute value of one plus zero squared. But this is just equal to the natural logarithm of one, which we know is equal to zero. So this also contains a factor of zero. Therefore, it’s equal to zero.

So our answer simplified to give us πœ‹ by four minus the natural logarithm of two divided by two. Therefore, by using integration by parts and what we know about definite integrals, we were able to show the integral from zero to one of the inverse tan of π‘₯ with respect to π‘₯ is equal to πœ‹ by four minus the natural logarithm of two divided by two.

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