u-substitution problem

**tabularasa** · November 14th 2008, 03:40 AM

Hi,

I want to use u-substitution in my self-study-case.

If I have equation: $\int (x^2-1)^2\:dx$ and then I substitute:

$let\:\:u = (x^2-1)$

$du \:=\: 2x \:dx$

$dx\:=\:\frac{du}{2x}$

And then I plugin the u:
$\int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\:$

And finally plugin the original x:
$\frac{1}{2x}\:\times\frac{(x^2-1)^3}{3}\:$

Whats wrong with my substitution. I want to use $let\:\:u = (x^2-1)$ Is this substitution even possible?

**jhomie** · November 14th 2008, 04:00 AM

I dont know if substitution would work here, but not 100% sure. I know you could easily open up the brackets to find the integral to be: $\frac {x^5}{5} - \frac {2x^3}{3} + x$

**tabularasa** · November 14th 2008, 04:33 AM

Originally Posted by jhomie

I dont know if substitution would work here, but not 100% sure. I know you could easily open up the brackets to find the integral to be: $\frac {x^5}{5} - \frac {2x^3}{3} + x$

Yes that is true. I still looking solution for the substitution whether it's possible or not.

**HallsofIvy** · November 14th 2008, 04:47 AM

Originally Posted by tabularasa

Yes that is true. I still looking solution for the substitution whether it's possible or not.

No it is not possible. "Substitution" is essentially the reverse of "chain rule" in differentiation:
If f is a function of u and u is a function of x, then df/dx= (df/du)(du/dx)
When differentiating, we are free to just go ahead and multiply that "du/dx" in.

If we have an integral $\int f(u) (du/dx)dx$ , that is the same as [tex]\inf f(u)du[/itex]. Unfortunately, if that "du/dx" is not inside the integral to begin with, we cannot just put it in: we cannot, for example, say "if $u= x^2-1$ so that $du= 2x dx$ so $\int (x^2- 1)^2 dx= \frac{1}{2x}\int (x^2-1)^2 (2xdx)= \frac{1}{2x}\int u^2 du$ because you cannot cancel an x outside the integral with an x inside the integral.

You CAN cancel constants inside and outside the integral because [tex]\int c f(x)dx= c\int f(x)dx[/itex] for c a constant. If that problem were $\int (3x-1)^2 dx$ , then we could say "let u= 3x- 1 so du= 3dx. Then $\int (3x-1)^2 dx= \frac{1}{3}int (3x-1)^2 (3dx)= \frac{1}{3}\int u^2 du$ .

But in general we cannot make the subsitution u= f(x) unless du/dx is alread inside the integral. That's one thing that make integration so much harder than differentiation.

**albi** · November 14th 2008, 04:48 AM

You cannot do that:
$ \int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\: $

You should not treat x as a constat, because it depends on u. The only way to do that would be to express x in a terms of u. You can do that solving the equation for x:
$x^2 - (u+1) = 0$ since u+1 >0 we get

$x = \pm \sqrt{u+1}$

Hence the integral is
$ \frac{1}{2}\int u^2\: \frac{du}{\pm \sqrt{u+1}}$

Good luck.

**tabularasa** · November 14th 2008, 05:23 AM

Originally Posted by albi

You cannot do that:
$ \int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\: $

You should not treat x as a constat, because it depends on u. The only way to do that would be to express x in a terms of u. You can do that solving the equation for x:
$x^2 - (u+1) = 0$ since u+1 >0 we get

$x = \pm \sqrt{u+1}$

Hence the integral is
$ \frac{1}{2}\int u^2\: \frac{du}{\pm \sqrt{u+1}}$

Good luck.

Thank you HallsofIvy and albi! This will help me a lot...I think

. Self-studying contines...

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