How can you evaluate the integral of \sqrt{R^2 - x^2} using substitution?

[soopo]

Homework Statement

How can you integrate the following?

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

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.

 

[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

Note from your diagram:
\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!

 

[soopo]

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

Note from your diagram:
\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!

You are right.
I get

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

 

[miqbal]

You evaluated

\int x \, dx

Whereas your original integral was

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

 

[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.

 

[soopo]

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.

I get this

\frac{\sin^{-1} \left( \frac{x}{R} \right) + \frac{\sin \left( {2<br /> \sin^{-1} \left( \frac{x}{R} \right)} \right)}{2}}{2}

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