1. ## u-substitution problem

Hi,

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

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

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

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

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

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

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

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

2. 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: $\displaystyle \frac {x^5}{5} - \frac {2x^3}{3} + x$

3. 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: $\displaystyle \frac {x^5}{5} - \frac {2x^3}{3} + x$
Yes that is true. I still looking solution for the substitution whether it's possible or not.

4. 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 $\displaystyle \int f(u) (du/dx)dx$, 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 $\displaystyle u= x^2-1$ so that $\displaystyle du= 2x dx$ so $\displaystyle \int (x^2- 1)^2 dx= \frac{1}{2x}\int (x^2-1)^2 (2xdx)= \frac{1}{2x}\int u^2 du$ 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 $\displaystyle \int (3x-1)^2 dx$, then we could say "let u= 3x- 1 so du= 3dx. Then $\displaystyle \int (3x-1)^2 dx= \frac{1}{3}int (3x-1)^2 (3dx)= \frac{1}{3}\int u^2 du$.

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.

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

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:
$\displaystyle x^2 - (u+1) = 0$ since u+1 >0 we get

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

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

Good luck.

6. Originally Posted by albi
You cannot do that:
$\displaystyle \int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\:$

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:
$\displaystyle x^2 - (u+1) = 0$ since u+1 >0 we get

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

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

Good luck.

Thank you HallsofIvy and albi! This will help me a lot...I think . Self-studying contines...