### Video Transcript

Find the most general
antiderivative capital πΊ of π£ of the function π of π£ equals four cos π£ plus
three over five root one minus π£ squared.

Remember, the antiderivative is
basically the opposite of the derivative. And another way of thinking about
it is to find the antiderivative capital πΊ of π£. We can find the indefinite integral
of this function. We, therefore, say that capital πΊ
of π£ is equal to the indefinite integral of lower case π of π£. Letβs replace π of π£ with the
function four cos π£ plus three over five times the square root of one minus π£
squared. We then recall a key property of
integrals; that is, the integral of the sum of two or more functions is equal to the
sum of the integral of each respective function. And we can, therefore, split our
integral up. And we see that capital πΊ of π£ is
equal to the integral of four cos π£ plus the integral of three over five times the
square root of one minus π£ squared.

Another key property we can apply
is that the integral of some constant times a function is equal to that constant
times the integral of the function. And so we can further rewrite this
as four times the integral of cos of π£ plus three-fifths times the integral of one
over the square of one minus π£ squared. Now, this is great because we can
use in general results for derivatives. Firstly, we know that the
derivative of sin π₯ is cos of π₯. So the antiderivative and,
therefore, the integral of cos of π£ is sin of π£. And of course, when dealing with
definite integrals, we add a constant of integration. Letβs call that π΄. So this first part becomes four
times sin of π£ plus π΄.

Next, we know that if the inverse
sin of π₯ over π is greater than or equal to negative π by two and less than or
equal to π by two, then its derivative is equal to one over the square root of π
squared minus π₯ squared. Now, in our example, π squared is
equal to one. So π must be equal to one
also. So the antiderivative of one over
the square root of one minus π£ squared and, therefore, the integral of this
function is the inverse sin of π₯ over one. And we add another constant of
integration π΅. Now, of course, the inverse sin of
π₯ over one can be written as the inverse sin of π₯.

Distributing our parentheses and
combining the constants into one new constant capital πΆ, we find that the general
antiderivative capital πΊ of π£ is equal to four sin of π£ plus three-fifths of the
inverse sin of π£ plus πΆ. And in this example, weβve seen
that the indefinite integral of one over the square root of π squared minus π₯
squared with respect to π₯ is the inverse sin of π₯ over π plus π.