1. Dec 18, 2004

### bjon-07

Intergrate (sin(x^(1/2))

The problems tell me to do a subsition then use by parts to solve it.

2. Dec 18, 2004

### arildno

Set $$u=\sqrt{x}$$

3. Dec 18, 2004

### mathwonk

guessing and playing around is also a useful technique sometimes. just looking at that i would try xsin(x^(1/2)) just to see what comes out. well i got some stuff with x^(1/2) in front, which led me to try x^(1/2)cos(x^(1/2)). that came close enough to guess the rest.

guessing and playing around is actually easier and faster sometiems than keeping track of all the products and signs in the "parts" algorithm.

4. Dec 18, 2004

### JasonRox

When a question refers to certain style, i.e. "substitution", it is best to read the chapter to find out what it meant by it or atleast look it up in the textbook.

I recommend reading the sections of the chapters you have trouble with. If you don't read the text at all, may I ask why did you buy it?

5. Dec 18, 2004

### Benny

With questions which tell you to use a certain method I think it is best to give it a go even if you do not see where it might lead.

$$\int {\sin \left( {\sqrt x } \right)} dx$$

Let $$u = \sqrt x \Rightarrow \frac{{du}}{{dx}} = \frac{1}{{2\sqrt x }} \Rightarrow dx = 2\sqrt x du$$

From the substitution made earlier you can write: $$dx = 2udu$$

So you now have:

$$\int {\sin \left( {\sqrt x } \right)} dx$$

$$= \int {\sin \left( u \right)2u} du$$

$$= - 2u\cos \left( u \right) - \int {\left( { - \cos \left( u \right)} \right)} 2du$$

$$= - 2u\cos \left( u \right) + \int {2\cos \left( u \right)} du$$

$$= - 2u\cos \left( u \right) + 2\sin \left( u \right) + c$$

$$= - 2\sqrt x \cos \left( {\sqrt x } \right) + 2\sin \left( {\sqrt x } \right) + c$$

6. Dec 18, 2004

### bjon-07

Thanks alot for the help. Its all comming back to me now. I completey forgot that you can have a U in your du expression.