# Integration by parts

1. Dec 11, 2005

### Yann

I've got a simple, at least it seems so;

$$\int \sqrt{9-x^2}dx$$

I MUST solve it "by parts" (withtout trigonometric substitutions), but i'm stuck. If i choose u = (9-x^2)^(1/2), du = -x/((9-x^2)^(1/2)), dv = dx, v = x. I then have;

$$x\sqrt{9-x^2} + \int \frac{x^2dx}{\sqrt{9-x^2}}$$

Everything I do just make the equation more complicated. I know I have to choose another u and dv, but every choice I made only make it worse.

2. Dec 11, 2005

### emptymaximum

the thing to remember about integration by parts is that in the first integral you need to have a PRODUCT of two functions.
$$\int u dv = u v - \int v du$$
so the real question is, how do you make $\sqrt{9 - x^2}$ be a product of two terms?

Last edited: Dec 11, 2005
3. Dec 11, 2005

### Yann

It's common to use "dx" as "dv", it doesn't seem to be helping me on this one though. And $$\sqrt{9-x^2}$$ can be a product, but it's not necessarily something easy to integrate and that's my problem, I can't get it only integrable parts.

4. Dec 11, 2005

### emptymaximum

no, you can't use dx = dv. that isn't going to help you out, because you'll need to integrate x in the next step and the more steps you have, the more you integrate x and you can do that forever. why?
in order to use integration by parts, the integrand needs to be a product of two functions of x. here is why this is so:
using the standard notation of capital being anti-derivative (integral f(x) dx = F(x))
$$\int U(x) v(x) dx = U(x) V(x) - \int V(x) u(x) dx$$
if you like prime notation better (i do so i'm gonna use it ):
$$\int u(x) v'(x) dx = u(x) v(x) - \int v(x) u'(x) dx$$
this shows really good what integration by parts is doing. if you rearange the terms you get:
$$u(x) v(x) = \int u(x) v'(x) dx + \int u'(x) v(x) dx$$
if you take the derivative wrt x on both sides you get:
$$\frac{d} {dx} ( u(x) v(x) ) = u(x) v'(x) + u'(x) v(x)$$
look framiliar? it should, it's the PRODUCT rule.
So, in order for integration by parts to work, the integrand in the integral you are trying to solve needs to be a product of 2 functions. the integrand you have been given is:
$$\sqrt{9 - x^2} = f(x)$$
what you need to do is figure out how to express f(x) as a product of two functions.
$$f(x) = u(x) v(x)$$
choosing v'(x) = 1 isn't going to get the job done seeing as how it isn't a function of x, it is a constant (1). It is just going to add another term to the integral on the RHS and what you're trying to do is simplify the integral, not add extra terms to it. (If you can pick a u such that du = 1 , that's great, the integral on the RHS will be that much easier.)

how can you express $\sqrt{9 - x^2}$ as a product of two FUNCTIONS of x?

Last edited: Dec 11, 2005
5. Dec 11, 2005

### Yann

It can be usefull to take 1 as dv (for the integral of ln(x) for example), but you're right it's useless here. However, the problem is that (9-x^2) are fused together, I can't get them out unless i make it 9/(9-x^2)^(1/2)-x^2/(9-x^2)^(1/2). And I can't use (9-x^2)^a*(9-x^2)^b, because I'd need to integrate at least one of them and I don't have the necessary -2x. So my problem is really choosing the right "u" and "dv".

6. Dec 11, 2005

### emptymaximum

why do you need to get them out?
you can represent (9-x²) as a product of two functions of x easily.
If f(x) is a second degree polynomial, then f(x) = g(x)h(x) by definition.
hint: try factoring f(x)

Last edited: Dec 11, 2005
7. Dec 12, 2005

### Yann

I'm really sorry, maybe it's stress, but I don't know how to do that kind of equation and exams are soon. I don't know how to put $$\sqrt{9-x^2}$$.

8. Dec 12, 2005

### Benny

You used parts on sqrt(9-x^2). For the integral that results from this technque you could try to make use of:

$$\frac{{x^2 }}{{\sqrt {9 - x^2 } }} = x\frac{x}{{\sqrt {9 - x^2 } }}$$

$$\frac{d}{{dx}}\left( {\sqrt {9 - x^2 } } \right) = - \frac{x}{{\sqrt {9 - x^2 } }}$$

Use the above to integrate by parts, the integral you are have trouble with. See what you come up with, it should be fairly easy to deduce the answer. Hope this helps.

9. Dec 12, 2005

### emptymaximum

9 - x² is a difference of squares;
9 - x² = (3-x)(3+x) = f(x)g(x)
then:
√(9 - x²) = √[(3-x)(3+x)] = √(3-x)√(3+x)

10. Dec 12, 2005

### VietDao29

???
WHAT??? What do you mean???
Why can't I choose dv = dx? What's wrong with that choice?
Suppose you are to integrate the function:
$$\int \ln x dx$$, how are you going to you tackle this problem without letting u = ln(x) and dv = dx???
Am I missing something?
----------------------
I am confused... How can I continue if I let u = √(3-x) and dv = √(3+x)dx?
Am I missing something here?
----------------------
@Yann: you should follow Benny's suggestion.

Last edited: Dec 12, 2005
11. Dec 12, 2005

### Yann

Let's say I take (3-x)^(1/2) as u, (3+x)^(1/2) as dv;

$$\frac{2}{3}(3-x)^{1/2}(3+x)^{3/2}-\int \frac{1}{2}(3-x)^{-1/2}\times\frac{2}{3}(3+x)^{3/2}$$

It's a dead end. Same thing If i take (9-x^2)^(1/2) as u and dx as dv;

$$x(9-x^2)^{1/2}-\int x\frac{-xdx}{(9-x^2)^{1/2}}$$

I really need to know how to solve it, I don't care about the answer I can have it I want to see how it can be solved so I won't have problem at my exam... I start to wonder, at least is it possible ??? I've ask lots of people, nobody ever found how !

12. Dec 12, 2005

### Jameson

Yes, since this involves a sine inverse in the answer, I don't see how this can be done without trig substitution. Have you consulted with your teacher?

13. Dec 12, 2005

### Yann

Well, I'll try with a substitution;

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

$$x = a\sin\theta$$

$$dx = a\cos\theta\theta$$

$$\theta = \sin^{-1}(x/a)$$

$$\int a\sqrt{\cos^2\theta}a\cos\theta d\theta$$

$$\int a^2\cos^2\theta d\theta$$

$$cos^2=\frac{1}{2}(1+\cos(2\theta))$$

So;

$$\int \frac{a^2}{2}+\frac{a^2}{2}\cos(2\theta)$$

$$\frac{a^2\theta}{2}+\frac{a^2}{4}\sin(2\theta) +C$$

$$\sin 2u = 2\sin\cos$$

$$\frac{a^2\theta}{2}+\frac{a^2}{2}\sin\theta\cos\theta +C$$

Getting rid of $$\theta$$...

$$\frac{a^2}{2}\sin^{-1}(x/a)+\frac{x}{2}\sqrt{a^2-x^2}+C$$

So, with a = 3;

$$\int \sqrt{9-x^2}dx=\frac{9}{2}\sin^{-1}(x/3)+\frac{x}{2}\sqrt{9-x^2}+C$$
Is that right ?

Last edited: Dec 12, 2005
14. Dec 13, 2005

### VietDao29

Yup. That looks good.
-------------------
By the way, have you covered that the derivative of Arcsin(x) is: $$\frac{1}{\sqrt{1 - x ^ 2}}$$? If you have already covered that, then you don't really need to use trigonometric substitutions.
Now:
$$\frac{d\arcsin (x)}{dx} = \frac{1}{\sqrt{1 - x ^ 2}}$$, that means:
$$\int \frac{dx}{\sqrt{1 - x ^ 2}} = \arcsin (x) + C$$, right?
Now we'll find the anti-derivative to the function:
$$\int \frac{dx}{\sqrt{a ^ 2 - x ^ 2}}$$. (a > 0). Let x = at => dx = a dt.
$$\int \frac{dx}{\sqrt{a ^ 2 - x ^ 2}} = \int \frac{a \ dt}{\sqrt{a ^ 2 - a ^ 2 t ^ 2}} = \int \frac{a \ dt}{a\sqrt{1 - t ^ 2}} = \int \frac{\ dt}{\sqrt{1 - t ^ 2}} = \arcsin (t) + C = \arcsin \left( \frac{x}{a} \right) + C$$.
----------------------
Now, go back to your problem, by integrating by parts, you have:
$$\sqrt{a ^ 2 - x ^ 2} \ dx = x \sqrt{a ^ 2 - x ^ 2} \ + \ \int \frac{x ^ 2 \ dx}{\sqrt{a ^ 2 - x ^ 2}} = x \sqrt{a ^ 2 - x ^ 2} \ + \ \int \frac{- (- x ^ 2 + a ^ 2) + a ^ 2}{\sqrt{a ^ 2 - x ^ 2}} \ dx$$
$$= x \sqrt{a ^ 2 - x ^ 2} \ - \ \int \sqrt{a ^ 2 - x ^ 2} dx \ + \ \int \frac{a ^ 2}{\sqrt{a ^ 2 - x ^ 2}} \ dx$$
Rearrang it a bit, we have:
$$2 \int \sqrt{a ^ 2 - x ^ 2} dx = x \sqrt{a ^ 2 - x ^ 2} \ + \ \int \frac{a ^ 2}{\sqrt{a ^ 2 - x ^ 2}} \ dx = x \sqrt{a ^ 2 - x ^ 2} \ + \ a ^ 2 \arcsin \left( \frac{x}{a} \right) + C$$
$$\Leftrightarrow \int \sqrt{a ^ 2 - x ^ 2} dx = \frac{x}{2} \sqrt{a ^ 2 - x ^ 2} \ + \ \frac{a ^ 2}{2} \arcsin \left( \frac{x}{a} \right) + C$$
What is exactly what you have by using trigonometric substitutions. Congratulations...

Last edited: Dec 13, 2005