# Changing form of integral

1. Jun 11, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

Use integration by parts to show that

∫(sqrt(1-x^2) dx = x(sqrt(1-x^2) + ∫ (x^2) / sqrt(1 - x^2)

write x^2 = x^2 -1 + 1 in the second integral and deduce the formula

∫(sqrt(1-x^2) dx = (1/2)x(sqrt(1-x^2) + (1/2)∫ 1 / sqrt(1 - x^2) dx

I actually found a solution guide to this problem online. Although I don't even understand the solution. I do but I'm stuck on something. Please see the pdf attached. It is exercise 3.

2. Relevant equations

3. The attempt at a solution

OK so I'm confused on how they got the first integral after the = sign which reads -∫sqrt(1-x^2) dx

So what I tried to do was what the book tells me to do.

I sub so I got ∫x^2 / sqrt(1-x^2)dx = ∫ (x^2 -1) / sqrt(1-x^2) dx + ∫ (1/ sqrt(1-x^2)) dx

So if you look the integral on the right is OK that takes care of the last integral in eq 2 in the attached pdf. But I'm stuck with this integral and I have tried literally everything I can think of to get it to the form -∫sqrt(1-x^2) dx from ∫ (x^2 -1) / sqrt(1-x^2) dx

I tried to let u^2 = 1 - x^2 and sub back and sub the numerator for x^2 in terms of u.
I tried to let u = 1 - x^2.
I tried to let u = the numerator and I tried u^2 = numerator. I have tried trig substitution for the denominator I tried to draw a triangle and make sqrt ( 1 - x ^2) the adjacent side and proceed from there. NOTHING! This is frustrating because I have tried everything I know and I guarantee I will be mad because it is probably something simple I can't see. I feel I have exhausted my efforts

Just look at the attached note for exercise 3. You will see.
Thanks

#### Attached Files:

• ###### caltech note.pdf
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2. Jun 11, 2013

### CAF123

So what you have is: $$\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx + \int \frac{1}{\sqrt{1-x^2}}\,dx$$

All they are doing to get the new form $-\int \sqrt{1-x^2}\,dx$ for the first term on the RHS is taking out a common factor on the numerator. What is that common factor?

3. Jun 11, 2013

### Jbreezy

$\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx + \int \frac{1}{\sqrt{1-x^2}}\,dx$

Just a factor from this?
$I = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx$

$I = \int \frac{(x-1)(x+1)}{\sqrt{1-x^2}}\,dx$

This is sad I don't see it.

4. Jun 11, 2013

### CAF123

Okay, I'll start you off: (You'll be kicking yourself)
$$x^2 - 1 = -(1-x^2)$$

Now simplify.

5. Jun 11, 2013

### Jbreezy

My foot hurts. Beyond dumb.