# Integration of (R^2 - x^2)^.5

1. Oct 28, 2009

### soopo

1. The problem statement, all variables and given/known data
How can you integrate the following?

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

3. The attempt at a solution

Integration by parts does not seem to be a solution here.

Integration by substitution works to some extend:

Let $x = R sin(u)$.

We get
$$\int \sqrt {R^2 - x^2} dx = \int Rcos(u) (R^2 + 2R^2 sin^2(u)) Rsin(u) du$$
$$= \int R^4 cos^2(u) sin(u) (1 + 2sin^2(u) ) du$$

I cannot see how to integrate this easily.
Perhaps, there is a better way to do this.

2. Oct 28, 2009

### miqbal

Seems like you've got the right idea. Solve this using trigonometric substitution:

1) Draw a right triangle. Label one of the vertices x, label the hypotenuse R.

Label the angle $$\vartheta$$ so $$sin(\vartheta)=x/R$$.

2) Substitution

$$x = Rsin(\vartheta), dx = Rcos(\vartheta)d\vartheta$$

$$\frac{\sqrt{R^{2} - x^{2}}}{R} = cos(\vartheta)$$

So $$\sqrt{R^{2} - x^{2}} = Rcos(\vartheta)$$

3) With that figured out your integral should be much simpler!

Last edited: Oct 28, 2009
3. Oct 28, 2009

### soopo

You are right.
I get

$$\frac{1}{2} R^2 \int sin(2\vartheta) = R^2 cos(2\vartheta)$$

4. Oct 28, 2009

### miqbal

You evaluated

$$\int x \, dx$$

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

5. Oct 28, 2009

### miqbal

You need to evaluate

$$\int R^{2}cos^{2}(\vartheta) \, d\vartheta$$

Make sure you after you solve the integral you substitute back for the solution in terms of x.

6. Oct 28, 2009

### soopo

I get this

$$\frac{\sin^{-1} \left( \frac{x}{R} \right) + \frac{\sin \left( {2 \sin^{-1} \left( \frac{x}{R} \right)} \right)}{2}}{2}$$

It should be correct, since SageMath gives the same result.