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

[chipotleaway]

Homework Statement

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

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

 

[Ray Vickson]

Homework Statement

Using the substitution x=acos^2(\theta)+bsin^2(\theta)

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

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

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

 

[chipotleaway]

Oh yeah, typo. Fixed it, thanks :p

 

[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

 

[Mark44]

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

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.

 

[chipotleaway]

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.

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

 

[haruspex]

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

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.

 

[chipotleaway]

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.

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.