# Evaluation trig sub

1. Nov 4, 2007

### Winzer

1. The problem statement, all variables and given/known data
Having trouble evaluating:
$$\int_{\infty}^{-\infty} \frac{dz}{(z^2+x^2)^(3/2)}$$

2. Relevant equations
Trig sub
$$z=xtan(\theta)$$

3. The attempt at a solution
comes down to:
$$\int_{x\frac{\pi}{2}}^{-x\frac{\pi}{2}} cos(\theta) d\theta$$
goes to
$$sin(\theta)$$ from $$x\frac{\pi}{2} \longrightarrow -x\frac{\pi}{2}$$
ehh!?

Last edited: Nov 4, 2007
2. Nov 4, 2007

### Gib Z

You can change the bounds differently. Think about it, as long x is some constant, and z approaching negative infinity and positive infinity, the tan theta has to be...

3. Nov 4, 2007

### Winzer

mm...just $$\frac{\pi}{2}$$

4. Nov 4, 2007

### rocomath

is that supposed to be to the 3/2 power or is that a constant multiplying the denominator?

5. Nov 4, 2007

### Winzer

Sorry it is that quantity raised to the (3/2)

6. Nov 4, 2007

### Winzer

Basically I am trying to figure out how this book got
$$\frac{z}{x^2\sqrt(z^2+x^2)}$$ with bounds from infinity to -infinity.
They then go on to get [1--1]=2.

I have to do this for a similar problem except my bounds will be L to -L.
the answer is: $$\frac{L}{x^2 \sqrt(L^2+x^2)}$$ I do not know how.
Trying to remember how to do bounds on trig subs.

7. Nov 4, 2007

Anyone?

8. Nov 4, 2007

### Antineutron

maybe divide top and bottom by z. then plug in the limits from infinity to -infinity

I get zero??? I get (1/x^2)-(1/X^2)

9. Nov 5, 2007

### Gib Z

Integrating with respect to z or x? State the exact question given to you please.