Video: Finding the General Antiderivative of a Given Function Involving Trigonometric Functions

Find the most general antiderivative capital 𝐺(𝑣) of the function 𝑔(𝑣) = 4 cos 𝑣 + 3/5√(1 βˆ’ 𝑣²).

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Video Transcript

Find the most general antiderivative capital 𝐺 of 𝑣 of the function 𝑔 of 𝑣 equals four cos 𝑣 plus three over five root one minus 𝑣 squared.

Remember, the antiderivative is basically the opposite of the derivative. And another way of thinking about it is to find the antiderivative capital 𝐺 of 𝑣. We can find the indefinite integral of this function. We, therefore, say that capital 𝐺 of 𝑣 is equal to the indefinite integral of lower case 𝑔 of 𝑣. Let’s replace 𝑔 of 𝑣 with the function four cos 𝑣 plus three over five times the square root of one minus 𝑣 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 𝐺 of 𝑣 is equal to the integral of four cos 𝑣 plus the integral of three over five times the square root of one minus 𝑣 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 𝑣 plus three-fifths times the integral of one over the square of one minus 𝑣 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 𝑣 is sin of 𝑣. And of course, when dealing with definite integrals, we add a constant of integration. Let’s call that 𝐴. So this first part becomes four times sin of 𝑣 plus 𝐴.

Next, we know that if the inverse sin of π‘₯ over π‘Ž is greater than or equal to negative πœ‹ by two and less than or equal to πœ‹ by two, then its derivative is equal to one over the square root of π‘Ž squared minus π‘₯ squared. Now, in our example, π‘Ž squared is equal to one. So π‘Ž must be equal to one also. So the antiderivative of one over the square root of one minus 𝑣 squared and, therefore, the integral of this function is the inverse sin of π‘₯ over one. And we add another constant of integration 𝐡. 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 𝐢, we find that the general antiderivative capital 𝐺 of 𝑣 is equal to four sin of 𝑣 plus three-fifths of the inverse sin of 𝑣 plus 𝐢. And in this example, we’ve seen that the indefinite integral of one over the square root of π‘Ž squared minus π‘₯ squared with respect to π‘₯ is the inverse sin of π‘₯ over π‘Ž plus 𝑐.

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