I Question about this Integration by Substitution

[Martyn Arthur]

TL;DR

Not able to understand final refernce to 1/6

This is part of the working from f(3x^2-1)^2xdx; I don't understand from when 6x becomes 1/6

 

[Mark44]

TL;DR Summary: Not able to understand final refernce to 1/6

This is part of the working from f(3x^2-1)^2xdx; I don't understand from when 6x becomes 1/6
View attachment 350019

The work shown above is very difficult to understand. Is this the integral you're trying to evaluate?
$$\int (3x^2 - 1)^2x~dx$$
If so, the obvious substitution is to let $u = 3x^2 - 1$, from which we get $du = 6x~dx$.

We can rewrite the original integral as $\frac 1 6\int (3x^2 - 1)^2~6xdx$

Note that we need to multiply xdx by 6 to get the expression for du, so we need to account for that by multiplying the integral by 1/6. IOW, we're multiplying everything by 6/6, or 1, which doesn't change the value of the integral.

Now, make the substitution with u and du, and you have an easy integral.

 

[Martyn Arthur]

Sorry it was vague, your deduction is correct.
I am revising this stuff after a while away.
I guess I am just missing something basic but I don't understand how the 1/6 is calculated; apologies in advance for being thick!

 

[Ibix]

You understand that if $u=3x^2-1$ then $du=6x\ dx$, right?

@Mark44 is multiplying the integral by $\frac 66$ (which is obviously 1, so changes nothing) but lets him do this:$$\begin{eqnarray}
&&\int(3x^2-1)^2x\ dx\\
&=&\frac 66\int(3x^2-1)^2x\ dx\\
&=&\frac 16\int(3x^2-1)^2\ 6x\ dx
\end{eqnarray}$$Then he can directly substitute the expressions in my first paragraph into (3) to get a really easy integral of $u$.

My preference is to rearrange $du=6x\ dx$ to $\frac 16du=x\ dx$ and substitute this and our definition of $u$ into (1). But that's entirely a matter of personal style, and you get the same answer either way.

 

[Mark44]

I guess I am just missing something basic but I don't understand how the 1/6 is calculated

Yes, you're missing something very basic. Integration by substitution is one of the simplest techniques for working with integrals. I'm hopeful that between my explanation and the one from @Ibix that you'll get the idea.

 

[Martyn Arthur]

You guys are very patient; thank you.
Ah I see I think.
It's not part of the actual technical substitution process but a mechanical process to get the integral into a figure that is convenient to work with.
If this is correct my thanks; there is not a need to take time responding further.
Martyn

 

[Mark44]

It's not part of the actual technical substitution process but a mechanical process to get the integral into a figure that is convenient to work with.

No, it actually is part of the substitution process. You start with an integral in terms of, say, x and dx. The substitution aims to get an integral in terms of, say, u and du, that is easier to evaluate.

 

[Martyn Arthur]

that is clear; many thanks
martyn