Evaluate the integral. ∫sin^2(πx) cos^5(πx) d x

Name: Problem 5, Chapter 7: Calculus Early Transcendentals
Uploaded: 2021-08-09T18:37:20-08:00
Duration: 8 min 29 s
Channel: Prakash Hampole
Description: Evaluate the integral. ∫sin^2(πx) cos^5(πx) d x

Key Concepts

Trigonometric Identities

Trigonometric identities, such as the Pythagorean identity (sin^2? + cos^2? = 1), are often used to rewrite expressions involving powers of sine and cosine. In integrals where one of the trigonometric functions is raised to an even power, these identities help convert the expression into a form that is easier to integrate.

Substitution Method

The substitution method, also known as u-substitution, is a key technique in integration. When integrating products of trigonometric functions like sine and cosine raised to various powers, one typically selects a substitution that simplifies the integral by setting u equal to one of the functions. The differential then naturally appears in part of the integrand, making the substitution effective.

Integration of Trigonometric Power Functions

Integrating trigonometric functions raised to powers often involves specific strategies depending on whether the exponents are even or odd. When one of the trigonometric functions has an odd power, it is practical to separate one instance of that function so that its derivative is present in the integral, while the remaining power is expressed in terms of the other function using trigonometric identities.

Transcript

00:01 So we're taking the integral of sine square of pi x times cost to fifth of pi x with respect to x.

00:08 And we're going to be using the pythagorean identity, which you probably know in the top right here.

00:14 And also the fact that the derivative of sine x is cost x.

00:18 Basically, when we have a sign to a power times a cost to a power in the integrand, if one of them is odd and the other is even, what we're going to do is we're going to look at the odd one.

00:31 In this case the cos and we're going to keep on taking out well in this case a coast squared until we're left with just one cos so that we can use this fact and use substitution now this will become a lot more clear when i actually start doing it so i'm going to start by taking out a coast squared from that coast to the fifth like this so this first part i can stay the same now i'm going to take out a coast squared and uh that leaves us with a coast cubed because 5 minus 3 would be sorry 5 minus 2 would be 3 and like i said we're going to do this keep doing this until we're left with just a cos you know with no exponents on it so i'm going to do that again basically to this coast cube so we have this first part and then i'm going to take out another coast squared from that coast cubed leaving us with just a cos pi just a cos pi x with no exponent on it.

01:58 So the point is these two parts, they're just cos squared.

02:03 So we can use this pythagin identity to rewrite them in terms of sign.

02:07 That way we'll just have a function of sign here and a derivative of sign here.

02:13 And we'll use use substitution.

02:15 So now i can use the identity, which of course shows that co squared x is equal to 1 minus sine squared x.

02:30 In this case, in place of x, we're just going to have pi x, of course.

02:34 So now i'm going to rewrite the two co -squareds, and we have this first part, and then in place of co -squared, we have 1 minus sine squared pi x, and then again, 1 minus sine squared pi x.

02:56 Squared pi x and finally coast pi x i'm just going to draw a line here to show that's not those aren't connected okay now all we have to do is u substitution because you see we just have a function of sine x here and we have cost pi x here which is a multiple of the of the derivative of sine pi x so what we're going to do is we're going to let uh you the sign pi x and that would mean that um d u would be equal to um pi time because of chain rule pi times kose of pi x and of course we don't actually have this pi here so we can just uh divide both sides by pi to get this um oh i'm sorry there should be a dx here but yeah we can just divide both sides of this by pi so that we get the coast pi x dx that we have here.

04:13 So of course that means basically 1 over pi d u is pi cos pi x no sorry it's a coast pi x dx so we can all that back into that integral that we had and we have um integral of sine squared pi no sorry in place of sign we're going to put u so we have u squared and then 1 minus u squared and then um another 1 minus u squared and then we have 1 over pi uh d u and we can always uh take we can always move the constant a constant term that's being multiplied outside of the integral so i'm just going to put that outside here...

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