How Does Substitution Simplify the Integration of \(3x^2(x^3 - 2)^4 dx\)?

[PlasmaSphere]

1. Find, by substitution, the integral of; 3x²(x³ - 2)⁴ dx
2. susbt'
3. u = x³ - 2, so du/dx = 3x², and du = 3x² dx

Now this is where I'm not sure what to do. As u = x³ - 2 you know that x = (u + 3)^1/3, and so i think you can write the integral as;

\int(u+3)^1/3.u⁴ du ... but i when i look at the answer, it seems that i don't need the (u+3)^1/3 part, as the answer is;

(1/5)(x³ - 2)⁵.

which i noted is just \intu⁴ duBut I can't see what happens to the 3x² from the original equation, as i would seem to get that same answer by integrating even if it wasn't there.

 

[neutrino]

Does rewriting the integrand as (x³-2)⁴3x² make it any easier?

 

[rocomath]

After the derivative of your substitution appears in your Integrand, you do not need to do another substitution.

\int3x^2(x^3-2)^4dx

u=x^3-2
du=3x^2dx

3x^2 is in your Integrand, that's it! You do not have a left over x term which requires you to manipulate your u-sub. to attain a x=(u+2)^{\frac 1 3} so that your Integrand is in terms of one-variable.

 

[PlasmaSphere]

I still don't get it fully, by follwing the examples in my book i seem to get a different result.
I'm going to follow the steps in my book of an example they gave, but put in this equation to the example. The original example is another one where x is the derivative of u, so it should be the same process as this. So i'll put in this equation to the example;

\int3x^2(x^3-2)^4dx

let u=(x^3-2), then;

du/dx=3x^2

now make that = xdx, as you only need xdx in the integral, not du=3x^2dx

which gives; (1/3)^{(1/2)}du=xdx

so \int3x^2(x^3-2)^4dx = \int(x^3-2)^4xdx

= \int(u)^4*(1/3)^{(1/2)}du

there's something wrong with that, i can't see what though

 

[rocomath]

How can you make that xdx? When the derivative of your u-sub is 3x^2dx?

\int3x^2(x^3-2)^4dx

u=x^3-2
du=3x^2dx

Last step before Integrating ...

\int u^4du

That's it.

Let's change up your original problem b/c I see why you're confused ... say your Integral is

\int x^3(x^3-2)^4dx

u=x^3-2
du=3x^2dx \rightarrow \frac 1 3 du=x^2dx

Well now you need divide that 3 to act as a constant (we'll come back to this later).

\int x (x^3-2)^4 x^2dx

Notice that your u-sub can only take care of the x^2 term while you will have an extra x term which requires you to manipulate your u-sub to get rid of it.

\frac 1 3 \int xu^4du

Going back to our u-sub, we must manipulate it so that we get x=u to replace x.

u=x^3-2 \rightarrow x=(u+2)^{\frac 1 3}

Now re-sub for your last x-term so that your Integral is in terms of only one-variable.

\frac 1 3 \int u^4(u+2)^{\frac 1 3}du

 

[Gib Z]

It sounds too obvious to be stating, but remember when a is equal to b, we can replace the letter b with a, and a with b, when we want to.

So for this;

\int (x^3-2)^4 3x^2 dx
We know that if we let u= (x^3-2), then du = 3x^2 dx. I'm sure you agree with us up to here.

Now, I have no idea what your thought process is, but we know that the (3x^2 dx) part is equal to du. Hence, we can write it as du instead of (3x^2 dx).

We also know the (x^3-2)^4 part is equal to u^4, so we write it as u^4 instead of what it is now. So we get
\int u^4 du. Simple as that.