# Integral Substitution

1. May 28, 2005

### cogs24

hi guys
im debating whether this question requires trignometric substitution or just normal substitution.

∫ √9-2(x-1)²

Im leaning towards normal substitution, with u = x-1, but im not sure
Any ideas
Thanx heaps

2. May 28, 2005

### whozum

Is it

$$\int \sqrt{9-(2x-1)^2}$$?

If so a normal one wont do, youll need a trig one.

Last edited: May 28, 2005
3. May 28, 2005

### dextercioby

Try to bring it to the form

$$C\int\sqrt{1-(something)^2} \ dx$$

,where C is a constant.

Daniel.

4. May 28, 2005

### cogs24

2 is on the outside of the (x+1)

5. May 28, 2005

### cogs24

(x-1) sorry

6. May 28, 2005

### Jameson

Well if you substitute u = x-1, you get the integral...

$$\int \sqrt{9 - 2u^2}du$$

I think trig substitution is where this is leading.

7. May 28, 2005

### GCT

yeah from what I remember, trig substitution was the way to go.

$$\sqrt(2) \int \sqrt{(9/2-u^{2})} ~du$$

Last edited: May 28, 2005
8. May 29, 2005

### cogs24

hi guys
im still stuck on this question, it has really got me stumped
i know that for the form

√a² - x².dx

x=asinθ, dx = acosθ.dθ

but i just cant seem to put it together

9. May 29, 2005

### dextercioby

Why not.U've got an |a| coming out of the sqrt and u'll get a square

$$\int \sqrt{a^{2}-x^{2}} \ dx =a|a|\int \cos^{2}\theta \ d\theta$$

Then use the double angle formula.

Daniel.

10. May 29, 2005

### cogs24

the 2 outside the (x-1)^2 is confusing me
Im not sure what my x and a squared is?

11. May 29, 2005

### dextercioby

Look at post #6.If you make the substitution

$$\sqrt{2}u=t$$,

u can put that integral under the form

$$\frac{1}{\sqrt{2}} \int \sqrt{3^{2}-t^{2}} \ dt$$

And now do that substitution involving "sin".

Daniel.

12. May 29, 2005

### cogs24

so, sqrt 2u = sin theta

13. May 29, 2005

### dextercioby

Nope.$u\sqrt{2}=3\sin\theta$

Daniel.