# Video: Finding the Expression of a Function given Its Derivative Using Indefinite Integration

Suppose that dπ¦/dπ₯ = 8 cscΒ² π₯ and π¦ = β12 when π₯ = π/3. Find π¦ in terms of π₯.

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### Video Transcript

Suppose that dπ¦ by dπ₯ is equal to eight times the csc squared of π₯ and π¦ is equal to negative 12 when π₯ is equal to π by three. Find π¦ in terms of π₯.

In this question, we want to find π¦ in terms of π₯. And to do this, weβre told that the derivative of π¦ with respect to π₯ is equal to eight times the csc squared of π₯. And this is a simple type of equation. If weβre told dπ¦ by dπ₯ is equal to some function of π₯, we call this a simple differential equation. And the reason we call this a simple differential equation is we can solve this by integrating both sides with respect to π₯. We get the integral of dπ¦ by dπ₯ with respect to π₯ is equal to the integral of π of π₯ with respect to π₯. But remember, integration and differentiation are opposite operations. So the integral of dπ¦ by dπ₯ with respect to π₯ is equal to π¦.

And assuming we can integrate our function lowercase π of π₯, letβs say this is equal to capital πΉ of π₯, then we can get an expression for π¦ in terms of π₯ up to a constant of integration πΆ. The question then becomes, how do we find our value of πΆ? And we can do this by using the fact when π¦ is equal to negative 12, π₯ is equal to π by three. So letβs start by integrating both sides of our simple differential equation. We get π¦ is equal to the integral of eight times the csc squared of π₯ with respect to π₯. And we know how to integrate the csc squared of π₯ with respect to π₯.

We recall for any constant π, the integral of π times the csc squared of π₯ with respect to π₯ is equal to negative π times the cot of π₯ plus a constant of integration πΆ. And this actually comes from the fact that the derivative of the cot of π₯ is equal to negative the csc squared of π₯. And we can then use the fact that differentiation and integration are opposite processes. So applying this integral rule with π equal to π, weβve shown that π¦ is equal to negative eight times the cot of π₯ plus the constant of integration πΆ. We now want to find our value of πΆ. And we remember the question tells us when π¦ is equal to negative 12, π₯ is equal to π by three. So weβll substitute both of these values into our expression. This gives us negative 12 is equal to negative eight times the cot of π by three plus πΆ.

Next, we can simplify this expression. We remember, for any value of π₯, the cot of π₯ is the same as saying one divided by the tan of π₯. So using this, we get negative 12 is equal to negative eight divided by the tan of π by three plus πΆ. And the tan of π by three is one of our standard trigonometric results which we should commit to memory. Itβs equal to the square root of three. So weβll add eight over root three to both sides of this expression to make πΆ the subject of our equation. We get negative 12 plus eight over root three is equal to πΆ. And we can simplify this expression for πΆ by rationalizing the denominator. We multiply eight over root three by root three over root three. And this gives us that πΆ is equal to negative 12 plus eight root three over three.

All we need to do now is substitute this expression for πΆ into our equation for π¦. We get that π¦ is equal to negative eight times the cot of π₯ minus 12 plus eight root three over three. And this is our final answer. Therefore, weβve shown if dπ¦ by dπ₯ is equal to eight times the csc squared of π₯ and π¦ is equal to negative 12 when π₯ is equal to π by three. Then π¦ is equal to negative eight times the cot of π₯ minus 12 plus eight root three over three.