# U substitution

1. Dec 8, 2009

### Zhalfirin88

1. The problem statement, all variables and given/known data
Solve using u-substitution.

$$\int \frac{1}{sin x} dx$$

3. The attempt at a solution
I looked up the integral answer and I can see the connection of the csc, but I don't know how to get this to be du/u for the ln function.

2. Dec 8, 2009

### Count Iblis

That's the last thing you should do when solving problems as it defeats the purpose of learning via solving problems.

There is no need to get things right after just one substitution. You can do substitution after substituton as many times you like. So why not start with putting sin(x) = u and see what you get first?

3. Dec 8, 2009

### Zhalfirin88

I already did all that, it's just too much to list. I tried u = sin x at first but didn't get very far. Then I thought that since the sin is on the bottom I could try cos x since -sin is derivative of cos but didn't get very far again. Then I tried rewriting 1/sin x using trig identities but I couldn't find any that would help. (:

EDIT: I have a question though. When I was doing u = sin x, du = cos x dx. So would this be correct?

$$\int \frac{1}{u}cos x dx$$

$$csc x \int \frac{du}{u}$$

$$csc x \ln sin x$$

Last edited: Dec 8, 2009
4. Dec 8, 2009

### Count Iblis

Let's start again then. If you put

sin(x) = u

you get:

dx/sin(x) = du/u 1/sqrt(1-u^2)

(up to a plus/minus sign which can be ignored for the moment). Do you agree?

Edit, I see your latest edit now. You have to express cos(x) in terms of u.

5. Dec 8, 2009

### Zhalfirin88

I'm not following what you have written there, particularly I don't see where you're getting 1/sqrt(1-u^2) from. Isn't that a derivative of the arcsin?

6. Dec 8, 2009

### Count Iblis

If you put

sin(x) = u

then

x = arcsin(u)

and thus:

dx = dx/du du = 1/sqrt(1-u^2) du

You can also say that:

sin(x) = u ------->

d[sin(x)] = du -------->

cos(x) dx = du ---------->

dx = 1/cos(x) du = 1/sqrt[1-sin^2(x)] du = 1/sqrt(1-u^2) du

So, we have:

dx/sin(x) = 1/u 1/sqrt(1-u^2) du

7. Dec 8, 2009

### Zhalfirin88

Ok, what's the next step?

edit: would you integrate after that? it looks like you'd have the natural log of a mess :P

8. Dec 8, 2009

### Count Iblis

Put u = 1/v. The idea is that the du/u part remains the same (up to a sign)

du = -1/v^2 dv

1/u = v

du/u = -1/v dv

but

1/sqrt(1-u2)

becomes

1/sqrt(1-1/v^2)

so, you can bring the v in the denominator inside the square root and end up with something simple.

9. Dec 8, 2009

### Zhalfirin88

gotcha, thanks