# Using this substitution, show that this integral is equal to the RHS

1. May 27, 2013

### chipotleaway

1. The problem statement, all variables and given/known data
Using the substitution $x=acos^2(\theta)+bsin^2(\theta)$

$\int^{b}_{a} \sqrt{(x-a)(b-x)}dx = \frac{\pi}{8}(b-a)^2$

3. The attempt at a solution

After making the substitutions and doing all the algebra, I have

$\int^{\frac{\pi}{2}}_{0} sin(\theta)cos(\theta)(b-a)dx$, with $dx=2(b-a)sin(\theta)cos(\theta)d\theta$

Now since the above expression appears as the integrand, shouldn't the integral become $\frac{1}{2}\int^{\frac{\pi}{2}}_{0}d\theta$? I can only get the RHS if I have $2\int^{\frac{\pi}{2}}_{0} sin^2(\theta)cos^2(\theta)(b-a)^2d\theta$

Last edited: May 27, 2013
2. May 27, 2013

### Ray Vickson

Are you sure you have the right problem? Should it be $\int_a^b \sqrt{(x-a)(b-x)}\, dx?$

3. May 27, 2013

### chipotleaway

Oh yeah, typo. Fixed it, thanks :p

4. Jun 4, 2013

### chipotleaway

Solutions haven't been provided for this particular problem set for some reason so I'm still curious as to whether the method is correct :p

5. Jun 5, 2013

### Staff: Mentor

You shouldn't need a solution. They have given you what the left side of the equation is supposed to be equal to, provided you haven't made mistakes in your work.

6. Jun 5, 2013

### chipotleaway

I probably should've said I don't understand the method. I don't know why we keep both the substitution and the differential substitution in the integrand

7. Jun 5, 2013

### haruspex

You correctly have this:
$\int^{\frac{\pi}{2}}_{0} sin(\theta)cos(\theta)(b-a)dx$, where $dx=2(b-a)sin(\theta)cos(\theta)d\theta$
From there, I don't understand the difficulty. You use the second expression to replace dx in the integral and get exactly what you wanted.

8. Jun 5, 2013

### chipotleaway

Ah, I see where I went wrong now... usually in evaluating something like $\int cos(x)(1-sin^2(x))$, we let u=sin(x) and so du=cos(x)dx and since it's already under the integral, we write $\int 1-u^2 du$. I blindly thought it was the same for this one and didn't know why we didn't get rid of the integrand when making the substitution.