calculus

[bold shore]

cant seem to solve this

[wind ledge]

do you know how to integrate functions ?

[warped hornet]

The question guides you through it. Can you show us what you do understand?

[violet sun]

You have to integrate the function

[silent kayak]

Maybe make a trig substitution and then u-sub or IBP

(1-(cosx)^2)cosx is the same function

You get cosx-(cosx)^3

You can integrate the cosx

Easy

And you can do the (cosx)^3 as (1-(sinx)^2)cosx and then you get cosx-cosx(sinx)^2

And then let u=sinx

And then maybe you know it

[warped hornet]

You're making it far more complicated than it is. Also, that's not what trig substitution means. Trig substitution is when you replace a variable with a trig function. You're just using a trig identity.

This question is clearly from a point in the learning process where they haven't done u-substitution yet. It's outlining a way around it.

[warped hornet]

You just went full circle back to where you started with two extra terms that are cos x and -cos x

[short jetty]

To find the antiderivative F(x) of f(x) = sin²(x) cos(x), we can use the power rule and the chain rule in reverse.

First, let's consider the derivative of k sin(x), where k is a constant:

d/dx (k sin(x)) = k cos(x).

Now, let's rewrite f(x) as a product of two functions:

f(x) = sin²(x) cos(x) = (sin(x))^2 * cos(x).

To find the antiderivative, we can rewrite f(x) as:

f(x) = (sin(x))^2 * cos(x) = (sin(x))^2 * cos(x) * 1.

Now, let's consider the function (sin(x))^2. We can rewrite it as:

(sin(x))^2 = (1/2)(1 - cos(2x)).

let's substitute this expression back into f(x):

f(x) = (1/2)(1 - cos(2x)) * cos(x).

To find the antiderivative F(x), we can integrate each term separately:

F(x) = ∫ (1/2)(1 - cos(2x)) * cos(x) dx.

Expanding the expression, we get:

F(x) = (1/2)∫ (cos(x) - cos(x)cos(2x)) dx.

Using the power rule, we can integrate each term separately:

F(x) = (1/2)(∫ cos(x) dx - ∫ cos(x)cos(2x) dx).

The integral of cos(x) is sin(x):

F(x) = (1/2)(sin(x) - ∫ cos(x)cos(2x) dx).

let's focus on the second integral. We can use integration by parts, with u = cos(x) and dv = cos(2x) dx:

du = -sin(x) dx, v = (1/2)sin(2x).

Using the integration by parts formula:

∫ u dv = uv - ∫ v du.

∫ cos(x)cos(2x) dx = (1/2)cos(x)sin(2x) - ∫ (1/2)sin(2x)(-sin(x)) dx.

Simplifying, we get:

∫ cos(x)cos(2x) dx = (1/2)cos(x)sin(2x) + (1/2)∫ sin(2x)sin(x) dx.

Using the identity sin(A)sin(B) = (1/2)(cos(A-B) - cos(A+B)), we can rewrite the integral:

∫ sin(2x)sin(x) dx = (1/2)∫ (cos(x-x) - cos(x+x)) dx.

∫ sin(2x)sin(x) dx = (1/2)∫ (cos(0) - cos(3x)) dx.

∫ sin(2x)sin(x) dx = (1/2)∫ (1 - cos(3x)) dx.

Integrating, we get:

∫ sin(2x)sin(x) dx = (1/2)(x - (1/3)sin(3x)) + C1,

where C1 is the constant of integration.

Substituting this back into the expression for F(x), we have:

F(x) = (1/2)(sin(x) - (1/2)cos(x)sin(2x) - (1/2)(x - (1/3)sin(3x))) + C.

Simplifying, we get:

F(x) = (1/2)sin(x) - (1/4)cos(x)sin(2x) - (1/4)x + (1/6)sin(3x) + C,

where C is the constant of integration.

[short jetty]

I hope you get it.

[warped hornet]

Are you trolling? This isn't the place.

[silent kayak]

I realized that when I did it

[silent kayak]

My fault sorry

I’m kind of new to the topics

I got the answer tho when I just did sinx=u

and then the cosxdx=du

So it’s integral u^2 du I think

Which is simple and then you just use reverse power rule and substitute back in sinx for u and add the constant

[warped hornet]

I could tell that you were just making a mistake. It was UmairAbbas whose answer is completely insane.

[warped hornet]

Note that if you read the question, it heavily suggests that it's meant for people who haven't learned substitution yet, so I think you're still missing the point.

[silent kayak]

Yea the derivative thing which would just be a chain rule

[deft oar]

.close quiz deadline passed #chill message

1. the derivative of k sin(x)^3 is 3k sin(x)^2 cos(x)
2. the integral of 3k sin(x)^2 cos(x) is k sin(x)^3 + C
3. the problem asks for sin(x)^2 cos(x), so if you set k = 1/3,
4. the integral of 3(1/3) sin(x)^2 cos(x) is 1/3 sin(x)^3 + C
5. the integral of sin(x)^2 cos(x) is 1/3 sin(x)^3 + C

the identity sin(A)sin(B) = (1/2)(cos(A-B) - cos(A+B))
would mean sin(2x)sin(x) = cos(2x-x) - cos(2x+x) which is a mistake