# Integration via substitution

1. Aug 11, 2012

### Mathpower

1. The problem statement, all variables and given/known data
Integrate the following via substituion:
t = r(r2 -4)5[/sub]

3. The attempt at a solution

I tried numerous variations. I tried substituting in u = r^(2/5) -4r^(1/5) also u=r^2 -4 and u = r^(11/5) -4r^(1/5)

NONE of these substitutions worked because of the lone r which happens to the co-effiecient of (r2 -4)5[/sub]. NO LUCK.

2. Aug 11, 2012

### AGNuke

I believe you want this.
$$\int tdt=\int r(r^2-4)^5dr$$

By taking $u=r^2-4$, you can transform the problem as, $\because du=2rdr$
$$\int tdt=\frac{1}{2}\int u^5du$$

3. Aug 11, 2012

### Mathpower

Awesome thank you...
Is there a trick to substitution questions?
ie: what is the best way to approach such questions?

4. Aug 11, 2012

### micromass

Staff Emeritus
Practice and experience.

5. Aug 11, 2012

### Mathpower

Apart from practice and experience (I do realise they are the main things...but apart from them)?

Don't say nothing...surely there must be some trick?

6. Aug 11, 2012

### micromass

Staff Emeritus
Sure, there are many tricks!! Too many to list here. Usually, a good calculus book will give you all the tricks you need. And then you only need practice and experience to know which trick to use at which time.

7. Aug 11, 2012

### Mathpower

Would you please be so kind as to post just 2 or 3 here? Pleeeeeaaase...

8. Aug 11, 2012

### micromass

Staff Emeritus
For example, when you see in your integral something like $f(x)^n$, then it make sense to make a substitution f(x)=u.
For example

$$\int \sqrt{1-x}dx$$

is of this form with f(x)=1-x and n=1/2. So a susbtitution u=1-x works.

This does not always work. For example, if you had

$$\int (x^2-1)^5 dx$$

then a substition $u=x^2-1$ won't help you (try it!!).
Sometimes we're lucky and we have

$$\int x(x^2-1)^5 dx$$

To see when this substitution helps, you need to try it out and get experience.

Another one, if you have something like f(ax+b) in the integral, then trying u=ax+b is sometimes helpful. For example

$$\int e^{10x+5} dx$$

can be solved this way. Again, this does not always help.

A very useful trick is when you have an integral like

$$\int \frac{f^\prime(x)}{f(x)}dx$$

then a substitution like u=f(x) always helps. For example, this allows you to solve problems like

$$\int \frac{1}{x\log(x)}dx$$

9. Aug 11, 2012

### Mathpower

Ok thank you so much. You are awesome!
Hmmmm... is it just my computer or is tex not working?

10. Aug 11, 2012

### Mathpower

Hey, what is the warning for? what did I do wrong?

11. Aug 11, 2012

### micromass

Staff Emeritus

12. Aug 11, 2012

### AGNuke

I haven't learnt integration (formally in Mathematics, and by extension, Integration of higher functions, by parts, etc. I still kinda use anti-derivative way to do integration). So, these tricks helps in getting most of integration I generally come across in Physics.

For example, You can always try to see if you can club a certain factor, like in ∫f(x)g(x)dx, f(x)=k.g'(x). If that's the case, like in the question you mentioned, you can always substitute it.

u=g(x); du=g'(x).dx = du=f(x).dx / k

Our integral becomes k∫g(x).[f(x).dx / k] = k∫u.du

Just remember trying to differentiate the bigger function and if you get something similar to smaller function, use substitution.

Also, for polynomial function, try seeing if there is a big function, and a lone variable with one lower degree is clubbable with dx, substitution.

And, practice. Physics had me practiced enough.