Question Video: Evaluating the Indefinite Integral of a Vector-Valued Function Mathematics

Video Transcript

Calculate the integral of the cos of two 𝑡 𝐢 minus two times the sin of 𝑡 𝐣 plus one over one plus 𝑡 squared 𝐤 with respect to 𝑡.

We can see we’re asked to calculate the indefinite integral of a vector-valued function. And we know this is a vector-valued function because it contains the unit directional vectors 𝐢, 𝐣, and 𝐤. It’s worth pointing out sometimes our vectors will be written with underlined notation, bold notation, or half-arrow notation. In this video, we’ll use half-arrow notation for vectors, and we’ll use hat notation to represent the unit directional vectors. But it doesn’t matter which vector notation you prefer.

Now, to evaluate the indefinite integral of a vector-valued function, we need to evaluate the indefinite integral component-wise. In other words, we need to integrate each of our component functions. There is one thing worth noticing: we’re subtracting our second vector, so we have two options. We could either integrate two times the sin of 𝑡 with respect to 𝑡 and then multiply this by negative one. Or we could bring the negative one inside of our component function. Either method would work; it’s personal preference which you would prefer.

So let’s start by evaluating the indefinite integral of our first component function. That’s the integral of the cos of two 𝑡 with respect to 𝑡. To help us evaluate this integral, we need to recall one of our standard integral rules for trigonometric functions. For any real constant 𝑎 not equal to zero, the integral of the cos of 𝑎𝑡 with respect to 𝑡 is equal to the sin of 𝑎𝑡 divided by 𝑎 plus a constant of integration 𝐶. In this case, our value of 𝑎 is equal to two, so we substitute this in. We get the sin of two 𝑡 divided by two, and we add a constant of integration we’ll call 𝐶 one. So we found the definite integral of our first component function. We want to do the same with our second component function.

Remember, we have a choice about what to do with our negative one. In this case, we brought it into our integral. So we want to evaluate the integral of negative two times the sin of 𝑡 with respect to 𝑡. And to evaluate this, we need to recall another one of our standard trigonometric integral results. The integral of negative the sin of 𝑡 with respect to 𝑡 is equal to the cos of 𝑡 plus 𝐶. We can use this to evaluate our integral. First, we take the constant factor of two outside of our integral. Then, we can integrate negative the sin of 𝑡 with respect to 𝑡 to get the cos of 𝑡 plus a constant of integration. We can simplify all of this to give us two times the cos of 𝑡 plus a constant of integration 𝐶 two.

We now need to evaluate the integral of our third component function. That’s the integral of one divided by one plus 𝑡 squared with respect to 𝑡. And to do this, we need to recall one of our derivative results for inverse trigonometric functions. We know the derivative of the inverse tan of 𝑡 with respect to 𝑡 is equal to one divided by one plus 𝑡 squared. And what this tells us is the inverse tan of 𝑡 is an antiderivative of one divided by one plus 𝑡 squared, which is our integrand. So we can evaluate this integral as the inverse tan of 𝑡 and we add a constant of integration we’ll call 𝐶 three.

We’re now ready to evaluate the integral of the vector-valued function given to us in the question. First, recall we found the indefinite integral of our first component function to be the sin of two 𝑡 divided by two plus 𝐶 one. So when we integrate this component-wise, this will be our first component function. We’ll write this as one-half times the sin of two 𝑡 plus 𝐶 one multiplied by 𝐢. Next, when we integrated our second component function, we got two times the cos of 𝑡 plus 𝐶 two. And since we brought the negative one inside of our component function, we need to add this. This gives us two times the cos of 𝑡 plus 𝐶 two 𝐣.

Finally, when we evaluated the integral of our third component function, we got the inverse tan of 𝑡 plus 𝐶 three. So we then get the inverse tan of 𝑡 plus 𝐶 three 𝐤. And we could leave our answer like this. However, there’s one more piece of simplification we can do. Consider what would happen when we expand our parentheses. We would get 𝐶 one multiplied by 𝐢 plus 𝐶 two multiplied by 𝐣 plus 𝐶 three multiplied by 𝐤. But remember, 𝐶 one, 𝐶 two, and 𝐶 three are all constant. So 𝐶 one 𝐢 plus 𝐶 two 𝐣 plus 𝐶 three 𝐤 will be a constant vector. And we could just call this constant vector 𝐂. And doing this gives us our final answer.

Therefore, we were able to evaluate the integral of this vector-valued function component-wise. We got that its integral was equal to one-half times the sin of two 𝑡 𝐢 plus two times the cos of 𝑡 𝐣 plus the inverse tan of 𝑡 𝐤 plus 𝐂.