# Integration question

1. Feb 23, 2005

### ToxicBug

Could anyone evaluate this integral for me? I got it in an exam and didn't know how to do it.

$$\int\sqrt{x-x^2}$$

Last edited: Feb 23, 2005
2. Feb 23, 2005

### vincentchan

let u = x-1/2 and do the substitude,...

3. Feb 23, 2005

### ToxicBug

What do you mean? Can you evaluate it completely showing the steps?

4. Feb 23, 2005

### Jameson

Ok, Mathematica was definitely wrong. There's going to be a trig substitution in this integral somewhere.

Last edited: Feb 23, 2005
5. Feb 23, 2005

### ToxicBug

Mathematica is **** and I'm sure thats not a right answer.

6. Feb 23, 2005

### vincentchan

what do I mean? I thought I make it very clear already!
$$\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx$$

let $$u=x-\frac{1}{2}$$

It become:

$$\int\sqrt{(\frac{1}{2}-u)(\frac{1}{2}+u)}du$$

$$=\int\sqrt{\frac{1}{4}-u^2 }du$$

$$=\frac{u}{4}\sqrt{1-4u^2}+\frac{1}{8}sin^{-1}(2u)+C$$

$$=\frac{x-1/2}{4}\sqrt{1-4(x-1/2)^2}+\frac{1}{8}sin^{-1}(2x-1)+C$$

7. Feb 24, 2005

### ToxicBug

Sick, thanks.

8. Feb 25, 2005

### dextercioby

And one more thing:"Mathematica" is ALWAYS RIGHT...:grumpy:

Daniel.

9. Feb 25, 2005

### HallsofIvy

Staff Emeritus
Cute factoring.

What I would have done is "complete the square": x- x2= 1/4-1/4+ x-x2= 1/4-(x-1/2)2 which would have led me to the u= x- 1/2 substitution. Once I had it in the form $\sqrt{\frac{1}{4}- u^2}$ I would use a trig substitution.