Solve each of the integrals . Some integrals require...

Solve each of the integrals . Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.
$$
\int\left(3-x^{2}\right)^{3 / 2} d x
$$

Key Concepts

Trigonometric Substitution

This is a method used in integration to simplify integrals that involve square roots of quadratic expressions. By substituting a trigonometric function for the variable, the radical can be expressed in terms of sine, cosine, or another trigonometric function, making the integral easier to evaluate.

Integration of Powers of Quadratic Expressions

This concept involves techniques for integrating expressions where a quadratic expression in the variable is raised to a power. Such integrals often require specific strategies, including substitution methods or trigonometric substitution, to break down the complexity of the integrand.

Back-Substitution to Algebraic Form

After applying a substitution such as a trigonometric substitution to simplify an integral, it is important to convert the result back into the original variable. This process, called back-substitution, allows the final answer to be expressed as an algebraic function rather than in terms of the substituted variable.

Transcript

00:02 I'm going to use trig substitution, although it might not be the best way.

00:07 Because we have a minus sign and the constant is first, i want to use the sign substitution.

00:14 So i get 3 minus x squared to the 3 halves is 3 minus 3 sine squared data to the 3 halves, 3 times 1 minus sine squared data to the 3 halves, 3 times 1 minus sine squared data to the 3 halves, 3 cosine squared data.

00:37 Theta to the three halves.

00:42 So 3 to the three halves cosine cubed and then x equals the square of 3 sine of theta.

00:55 So d x is the square root of 3 cosine theta d theta.

01:02 So this integral turns into 3 to the 3 halves cosine cube theta times square to 3 cosine theta d theta.

01:16 3 to the 3 halves and the square root of 3 gives you 3 to the 4 halves or 3 squared, which is 9.

01:24 And then you have cosine 4th theta, d theta.

01:29 This is why i think there might have been a better way.

01:31 Maybe integration by parts somehow, because i've got to do this.

01:37 Now i've got to put in the identity 1 plus cosine 2 theta over 2 squared because it's to the 4th power.

01:47 So that gives me 9 4 integral 1 plus 2 cosine 2 theta plus the cosine squared of 2 theta d theta.

02:04 All right, before i integrate, i'm going to fix that.

02:07 So i have 9 4s integral 1 plus 2 cosine 2 theta plus 1 plus cosine of 4 theta over 2 d theta.

02:25 So now i have this one and this one -half, so that gives me three - halves, plus two cosine two -theta, plus one -half cosine of four -theta, d -theta.

02:42 So i integrate and i get three - halves theta plus if u is two -theta, then d -u is two -d -theta, so that's going to get used up.

02:52 So i get the sign of two -theta here.

02:56 Then on this one, if u is four -theta, i need a four.

02:59 So that puts a one fourth here.

03:01 So that gives me one half, sorry, one eighth, sine of fourth theta plus c.

03:12 Okay, so the problem here is the sign of two, the two problem, sign of two theta and the sign of four theta.

03:21 So i'm going to have to use some identities...