(Basic) Please give me advice on my u-substitution technique

[939]

Homework Statement

I learned integration, but I think somewhere my technique is flawed, because I consistently get certain questions right and certain questions wrong. I don't know why I started doing it this way, but I always find DX by dividing it by DU and substitute it back into the equation. Where can I improve and what am I doing wrong?

Homework Equations

Here are two equations. First one I get right, second one wrong.
1) integral of (1 - 2x)^4 2dx
2) integral of ((5)(3t - 6)^2

The Attempt at a Solution

1)
U = 1 - 2x
DU = -2 dx
DX = DU/-2

(U)^4 (2)(DU/-2)
U^4 (-1)
-1 integral of (1/5) U^5 + C
= -1/5 (1-2x)^5 + c
RIGHT

2)
U = 3t - 6
DU = 3 dx
dx = (du/3)

5(u)^-2 du/3

1/3 5(u)^-1
(1/3) (-5/1) (3t - 6)^-1
WRONG

?

 

[Simon Bridge]

integral of ((5)(3t - 6)^2

... too many brackets. From your later statements I'm guessing you want:

$$\int 5(3t-6)^{-2} dt$$

However: Your substitution seems fine - I think you need to check how to integrate negative powers.

 

[SteamKing]

Is the exponent 2 or -2?

 

[Simon Bridge]

Wait:

(1/3) (-5/1) (3t - 6)^-1

... looks here like the power has been integrated properly (if it is -2). There is a minus sign missing in the previous line.

 

[Robert1986]

I think that u-substitution is a technique that is used at times when it really isn't necessary. IMO, u-substitution is a good way to program a computer to do an integration, but it is usually not a good way to integrate (well, really we do u-substitution to find an anti-derivative, which we use to do integration.) So, how would I do this problem?

Let's try to find the anti-derivative of2(1-2x)^4. Now, we have something raised to the power of 4, and so a good guess at an anti-derivative would be to just write down:
(1/5)(1-2x)^5. Now, we want to see if this is correct (ie is the derivative of our first guess what we started out with.) Well, the derivative is -2(1-2x)^4. This is VERY close to what we started out with, in fact, it is a multiplicative constant away (what constant?) So, just divide our initial guess by this multiplicative constant and check it again, and we see that this is the correct anti-derivative.

Now, if you try to do it like this, after a while, it will become second nature. It might have seen confusing the way I described it, but I think if you write it out, it will make sense. The fact of the matter is that u-substitution can sometimes make things MUCH simpler, but, IMO, these cases are few and far between. So, use u-sub when you HAVE to, but it should be a last resort (and after practice, you will realize which integrals really "need" u-sub.)