### 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, weβre 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 weβre 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₯, letβs say this is equal to capital πΉ of π₯, then we
can get an expression for π¦ in terms of π₯ up to a constant of integration πΆ. The question then becomes, how do
we find our value of πΆ? And we can do this by using the
fact when π¦ is equal to negative 12, π₯ is equal to π by three. So letβs 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 πΆ. 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, weβve shown that π¦ is equal to negative eight times the cot of π₯
plus the constant of integration πΆ. We now want to find our value of
πΆ. And we remember the question tells
us when π¦ is equal to negative 12, π₯ is equal to π by three. So weβll 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 πΆ.

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

All we need to do now is substitute
this expression for πΆ 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, weβve 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.