# Homework Help: Integrate (x-x^2)^1/2

1. Jun 2, 2012

### Math Monster

1. The problem statement, all variables and given/known data
Integrate (x-x^2)^1/2 from 1/2 to 1.

I tried to use substitution with x=sinu, dx=cos du to get:
sinu(1-sinu)^1/2 cosu du
but no idea where to go from there!

Also tried integration by parts, but it didn't work!

Help!

2. Jun 2, 2012

### Dickfore

Perform the Euler substitution:
$$\left( x (1 - x) \right)^{\frac{1}{2}} = x \, t$$
$$x (1 - x) = x^2 \, t^2$$
$$1 - x = x \, t^2$$
$$x = \frac{1}{1 + t^2} \Rightarrow \left(x (1 - x) \right)^{\frac{1}{2}} = x \, t = \frac{t}{1+t^2}, \ dx = -\frac{2 \, t \, dt}{(1 + t^2)^2}$$
$$x = \frac{1}{2} \Rightarrow t = 1, \ x = 1 \Rightarrow t = 0$$
$$I = \int_{1}^{0}{\frac{t}{1 + t^2} \, \frac{-2 t}{(1 + t^2)^2} \, dt} = 2 \, \int_{0}^{1}{\frac{t^2}{(1 + t^2)^3} \, dt}$$

Try using integration by parts on this integral. What should your u and v be?

3. Jun 2, 2012

### Math Monster

That is a great help I think!

I have worked through what you did, and got:

u=t^2
du = 2t dt

dv = (1+t^2)^-3
v = (1+t^2)^-2 / -4t

so integral of udv =
((t^2)(1+t^2)^-2) / (-4t) evaluated 0-1
+ integral_0 ^1 (1+t^2)^(-2) / -2 dt (the -4t from v has cancelled with the 2t from du)

If this is right, I'm able to finish it off from here!

Thanks very much. I've never seen the Euler method before, if you can provide a bit more information how to use it in other examples I would be grateful.

Kind regards

Harriet

4. Jun 2, 2012

5. Jun 2, 2012

### Math Monster

To work out v i:

1. Increased the power of the bracket from -3 to -2.
2. Divide by the new power (-2)
3. Divide by the differential of the bracket (2t)

Could you explain where I went wrong? Thanks Harriet

6. Jun 2, 2012

### Dickfore

In step 3. Notice that "the differential of the bracket" depends on t, your dummy variable.

7. Jun 2, 2012

### Mentallic

If you went backwards and tried to differentiate v with respect to t, you'd have to use the quotient rule.

8. Jun 2, 2012

### Math Monster

Oh okay, I'm a bit lost. I thought you could introduce the dummy variable and then differentiate/integrate as normal as you changed limits etc. Could you explain how it would be different?

Thanks

9. Jun 2, 2012

### Dickfore

An alternative substitution is obtained by completing the square under the square root:
$$x - x^2 = -\left(x^2 - 2 \frac{1}{2} x + \frac{1}{4} - \frac{1}{4} \right) = \left(\frac{1}{2}\right)^2 - \left( x - \frac{1}{2} \right)^2$$
Then, try the substitution:
$$x - \frac{1}{2} = \frac{1}{2} \sin t$$

What are the new limits of integration? What does the integral transform to?

10. Jun 2, 2012

### Math Monster

I rearranged to find the integral becomes 1/2cost
with t= 0, t=pi/2 for the new limits?

Any good?

11. Jun 2, 2012

### Dickfore

The limits are correct. Where did 1/2 cos t come from?

12. Jun 2, 2012

### Math Monster

Once you complete the square, substitute x-1/2 = 1/2sint to get 1/4 - 1/4sin^2 t
Factorise to get 1/4 ( 1-sin^2 t) = 1/4 cos^2 t
As we are looking at sqrt (x-x^2), take sqrt of this to get 1/2 cost

Does that make sense?

13. Jun 2, 2012

### Dickfore

Yes. So, it came from simplifying the square root. How did you transform dx then?

14. Jun 2, 2012

### Math Monster

dx = 1/2 cos t ?

So integral should be 1/4 cos^2 t?

15. Jun 2, 2012

### Dickfore

Yes. Now, when faced with a square of a sine or cosine, use the double-angle formula:
$$\cos^2 t = \frac{1 + \cos 2 t}{2}$$
Then, your integral should become elementary.

16. Jun 3, 2012

### quackdesk

I agree with Dickfore :) I had an example like that on my last exam :)