# Is this valid when doing u substitution for integration?

1. May 23, 2014

### TitoSmooth

So I'm doing length of an arc in my calculus 1 class. After plugging everything in the arc length formula.

Now I have this complicated function to integrate. Square root of (16x^8+8x^4+1)/16x^4.

I took the denominator out of my square root and got 4x^2.

Now I take u=4x^2.
Du/2x =dx.

Now my question is are we allowed to take the u substitution of a value in the denominator and not have to write it as

1/what we are sub?

. The problem is even so no solution. But it seemed to follow correctly if my intuition was correct.

2. May 23, 2014

### SammyS

Staff Emeritus
If what you substitute for is in the denominator, then what you substitute in has to be in the same place, the denominator.

Have you tried factoring what's left under the radical ?

3. May 23, 2014

### Ray Vickson

Show us what you actually get, so we can better judge the issue. As it stands now I cannot figure out what is your difficulty.

4. May 23, 2014

### TitoSmooth

X= (1/3) y^3 +1/4y. Find arc length between 1 and 3.

X prime squared= (4y^4-1)/4y^2

Now plug into arc length formula.

L=sqrt of (1+x prime squared)

Now L= sqrt of

I uploaded a pic to make it easy. My sub is on the middle of the left page

#### Attached Files:

• ###### 20140523_054631.jpg
File size:
20 KB
Views:
72
5. May 23, 2014

### TitoSmooth

The U=y^2 came from the denominator. When I take the u sub of something in the denominator can I write it as a linear function. Or must I always make it a rational function.

6. May 23, 2014

### SteamKing

Staff Emeritus
If u = 4x^2, then du = ?

7. May 23, 2014

### TitoSmooth

I factored the 1/4 out and put it outside the integral sign. So now my denominator is x^2. N du/2x= X was this step wrong? My question still isn't being answered. I'm not asking for help getting a solution but rather regarding how u substitution works when substituting something in the denominator for u.

8. May 23, 2014

### TitoSmooth

I attached an image to show my work. Can somebody comment on out? Thanks

9. May 24, 2014

### SammyS

Staff Emeritus
I think it was answered in post #2.

If a quantity is in the denominator, & you make a substitution for that quantity, the substitution goes where the original quantity was -- in the denominator.

It is possible (rarely) for you make some very clever substitution, so that some factor in the numerator will cancel the the denominator.

That attachment is a bit hard to follow. In the Original Post, you have the numerator being: $\ \sqrt{16x^8+8x^4+1\,}\ .$

Then in your attachment the same expression appears with y rather than x, except for a change of sign on the 4th degree term. After that, you sub. u for y2 and get $\displaystyle \sqrt{16u^4+8u^2+1\,}\,,\$ this time with the + sign returning. At any rate, at this point you see that $\displaystyle\ \ 16u^4+8u^2+1\$ factors to $\displaystyle \ \ (4u^2+1)^2\ .$(Corrected) This allows the radical to be removed altogether. Do a similar thing -- factor -- the original integrand. It will make life much easier.

Added in Edit: It was too late last night when I posted this !

As benorin points out below: That should be $\displaystyle \ \ (4u^2+1)^2\,,\$ rather than $\displaystyle \ \ (4u^2+4u+1)^2\ .\$ (Corrected in the above.)

Of course, benorin was working with $\displaystyle \ \ 16x^8+8x^4+1\$ and not $\displaystyle \ \ 16u^4+8u^2+1\ .\$

Last edited: May 24, 2014
10. May 24, 2014

### benorin

(16x^8+8x^4+1)=(4x^4+1)^2

11. May 24, 2014

### SteamKing

Staff Emeritus
It's not wrong, just confusing. If you want u = x^2, then write u = x^2. The constants can be dealt with separately.

The trick with u-substitution is to obtain an algebraically equivalent integrand once the u-substitution has been made, so that:

$\int f(x)dx$ = $\int g(u) du$

which, in turn, implies that:

f(x) dx = g(u) du

Multiplying the expression above by the same arbitrary constant on both sides does not
change the equality of the expressions.

Ideally, the expression g(u) du is one which can be integrated with a combination of the
integration techniques or formulas you have already learned.