# Basic Integration by substitution

1. Jan 6, 2008

### 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 dont need the (u+3)1/3 part, as the answer is;

(1/5)(x3 - 2)5.

which i noted is just $$\int$$u4 du

But I cant 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.

Last edited: Jan 6, 2008
2. Jan 6, 2008

### neutrino

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

3. Jan 6, 2008

### 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.

Last edited: Jan 6, 2008
4. Jan 6, 2008

### PlasmaSphere

I still dont 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 $$\int3x^2(x^3-2)^4dx$$ = $$\int(x^3-2)^4xdx$$

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

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

5. Jan 6, 2008

### 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.

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

Last edited: Jan 6, 2008
6. Jan 6, 2008

### 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.

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