### Video Transcript

Determine the family of functions π for which π triple prime of π‘ equals three sin π‘ minus two.

What does the notation π triple prime of π‘ mean? Itβs the notation for the third derivative. It means the function π of π‘ has been differentiated three times to get π triple prime of π‘. So we need to do the reverse of differentiation or integration three times to get π of π‘. So letβs get started and find the first antiderivative of three sin π‘ minus two. In the same way that we differentiate term by term, we can integrate term by term. So weβre integrating three sin π‘ and two. It might be helpful to recall some useful trick derivatives. Sin π‘ differentiates to cos π‘. Cos π‘ differentiates to negative sin π‘. Negative sin π‘ differentiates to negative cos π‘. And negative cos π‘ differentiates to sin π‘. This is helpful because, for integration, the arrows go exactly the opposite way.

Letβs also remind ourselves of integration of other functions. If we have a function such as four π₯ squared plus three π₯ plus five, we can differentiate this term by term. We start with four π₯ squared and multiply the coefficient four by the power two to get the new coefficient of eight. And then, we reduce the power by one to get eight π₯. Three π₯ is three π₯ to the power of one. So this differentiates to three. Itβs going to be helpful to us at this point to note that three π₯ differentiates to three. And so three integrates to three π₯. Five is a constant. So it differentiates to zero.

So thatβs differentiation. But what about antidifferentiation or integration? Well, itβs the reverse of differentiation. Instead of multiplying the coefficient by the power and subtracting one from the power, we increase the power by one and divide the coefficient by that new power. Because constants differentiate to zero, we account for that by adding a constant to our antiderivative. We often represent the constant with πΆ. But really we can use any letter.

So going back to our question, we want the antiderivative of three sin π‘ minus two. And this will give us π double prime of π‘. If we check our diagram, we see that sin π‘ integrates to negative cos π‘. So three sin π‘ integrates to negative three cos π‘. And two integrates to two π‘ as we saw in our example. And we must include the constant of integration, πΆ. And now we integrate again to find π prime of π‘. Our diagram shows that negative cos π‘ integrates to negative sin π‘. So negative three cos π‘ integrates to negative three sin π‘. To integrate two π‘, we follow our rules for antidifferentiation. Increasing the power of π‘ by one gives us the new power of two. And then, we divide the coefficient two by the new power of two, which is just one. So this is just π‘ squared. Weβve seen how to integrate a constant. So to integrate πΆ, we just get πΆπ‘. This is in the same way that two integrated to two π‘, πΆ integrates to πΆπ‘.

And finally, because weβve integrated, we have to include a constant of integration. So this time letβs call it π·. So weβve got π prime of π‘. And we have to keep going because, remember, weβre trying to get π of π‘. From our diagram, negative sin π‘ integrates to cos π‘. So negative three sin π‘ integrates to three cos π‘. To integrate π‘ squared, we increase the power by one to get the new power of three. And then, we divide by the new power. So we get π‘ cubed over three.

To integrate πΆπ‘, we follow our rules for antidifferentiation again. We increase the power by one and divide by the new power to give us πΆπ‘ squared over two. However, πΆ is an unknown constant. So dividing it by two means itβs still an unknown constant. So we can just write πΆπ‘ squared. π· is a constant. So that integrates to π·π‘. And our final step is to add a constant of integration, which we can call πΈ this time. So now we have π of π‘.

This is a family of functions because thereβs lots of possibilities for the constants πΆ, π·, and πΈ. All the possibilities for πΆ, π·, and πΈ create this family of functions. For a final answer, we often write the term with the highest power at the beginning of the function. In this case, itβs π‘ cubed over three.

So our final answer is negative π‘ cubed over three plus three cos π‘ plus πΆπ‘ squared plus π·π‘ plus πΈ.