# Trig substitution problem

1. Feb 22, 2007

### Sympathy

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

Evaluate the following integrals or state that they diverge. Use proper notation.

Integral from 0 to 2 of (x+1)/Square root(4-x^2)

2. Relevant equations

3. The attempt at a solution

I just substituted x = 2sin(theta) thus dx = 2cos(theta)

I got to the point where it is .5(integral from 0 to 2 of 2tan(theta)) + .5(integral from 0 to 2 of sec(theta)).

I think theres an easier approach at the problem.

suggestions?

2. Feb 23, 2007

### neutrino

Partial fractions.

3. Feb 23, 2007

### gammamcc

I think trig subst. is a good way. The limits of integration will change since you went from x to theta.

4. Feb 23, 2007

### dextercioby

So instead of going from 0 to 2, they go from 0 to $\pi/2$.

5. Feb 23, 2007

### gammamcc

Ya, that's it. Why do people have to take all the fun out of a problem? Why not leave a hint as a hint? This one-up business is boring.

6. Feb 23, 2007

### Gib Z

$$\int\frac{x+1}{4-x^2} dx =\frac{3}{4} \int \frac{1}{2-x} dx - \frac{1}{4}\int \frac{1}{2+x} dx$$. Far too simple from there, the only reason I posted even this far was because I wanted to do it as well :)

7. Feb 23, 2007

### Gib Z

Shoot me in the face, i didnt see the sqrt...

8. Apr 20, 2007

### ahmed jamal

hello friend

solution

Integral from 0 to 2 of (x+1)/Square root(4-x^2) dx

let x=2sin(&)
dx=2cos(&)d&

=Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1)(2cos&)/2cos& d&

=Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1) d&

i think know it is easy to solve