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

Suppose that d𝑦/dπ‘₯ = 8 cscΒ² π‘₯ and 𝑦 = βˆ’12 when π‘₯ = πœ‹/3. Find 𝑦 in terms of π‘₯.

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Video Transcript

Suppose that d𝑦 by dπ‘₯ is equal to eight times the csc squared of π‘₯ and 𝑦 is equal to negative 12 when π‘₯ is equal to πœ‹ by three. Find 𝑦 in terms of π‘₯.

In this question, we want to find 𝑦 in terms of π‘₯. And to do this, we’re told that the derivative of 𝑦 with respect to π‘₯ is equal to eight times the csc squared of π‘₯. And this is a simple type of equation. If we’re told d𝑦 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 d𝑦 by dπ‘₯ with respect to π‘₯ is equal to the integral of 𝑓 of π‘₯ with respect to π‘₯. But remember, integration and differentiation are opposite operations. So the integral of d𝑦 by dπ‘₯ with respect to π‘₯ is equal to 𝑦.

And assuming we can integrate our function lowercase 𝑓 of π‘₯, let’s say this is equal to capital 𝐹 of π‘₯, then we can get an expression for 𝑦 in terms of π‘₯ up to a constant of integration 𝐢. The question then becomes, how do we find our value of 𝐢? And we can do this by using the fact when 𝑦 is equal to negative 12, π‘₯ is equal to πœ‹ by three. So let’s start by integrating both sides of our simple differential equation. We get 𝑦 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‘Ž, the integral of π‘Ž times the csc squared of π‘₯ with respect to π‘₯ is equal to negative π‘Ž times the cot of π‘₯ plus a constant of integration 𝐢. 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‘Ž equal to π‘Ž, we’ve shown that 𝑦 is equal to negative eight times the cot of π‘₯ plus the constant of integration 𝐢. We now want to find our value of 𝐢. And we remember the question tells us when 𝑦 is equal to negative 12, π‘₯ is equal to πœ‹ by three. So we’ll substitute both of these values into our expression. This gives us negative 12 is equal to negative eight times the cot of πœ‹ by three plus 𝐢.

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œ‹ by three plus 𝐢. And the tan of πœ‹ by three is one of our standard trigonometric results which we should commit to memory. It’s equal to the square root of three. So we’ll add eight over root three to both sides of this expression to make 𝐢 the subject of our equation. We get negative 12 plus eight over root three is equal to 𝐢. And we can simplify this expression for 𝐢 by rationalizing the denominator. We multiply eight over root three by root three over root three. And this gives us that 𝐢 is equal to negative 12 plus eight root three over three.

All we need to do now is substitute this expression for 𝐢 into our equation for 𝑦. We get that 𝑦 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, we’ve shown if d𝑦 by dπ‘₯ is equal to eight times the csc squared of π‘₯ and 𝑦 is equal to negative 12 when π‘₯ is equal to πœ‹ by three. Then 𝑦 is equal to negative eight times the cot of π‘₯ minus 12 plus eight root three over three.

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