# Trig sub.

1. Feb 21, 2005

can someone help me find a appropriate trig sub for this problem:

$$\int\frac{x}{sqrt(-29-4x^2-24x)}$$

took out sqrt(4)...

$$sqrt(4)*sqrt(-29/4-x^2-6x)$$
(i also changed all the negative signs to positive)
complete the square....

$$sqrt(4)*sqrt((x+3)^2-7/4)$$

so my trig sub should be sqrt(7/4)*sec(t)-3, but it's incorrect

2. Feb 21, 2005

### ehild

Is it $$\int\frac{xdx}{sqrt(-29-4x^2-24x)}$$ or

$$\int\frac{dx}{sqrt(-29-4x^2-24x)}$$?

Try $$x=\sqrt{7/4}*\sin(t)-3$$

ehild

3. Feb 21, 2005

### dextercioby

It doesn't work with "sin",but it works with $\cosh t$...

Daniel.

4. Feb 21, 2005

### ehild

$$-29-4x^2-24x = 7-(2x+6)^2=7(1-(\frac{2x+6}{7})^2)$$
If $(2x+6)/\sqrt{7} \le 1$ then it can be substituted either by sin(t) or cos(t). In the opposite case, the integrand is imaginary and your substitution is OK.

ehild

5. Feb 21, 2005

### dextercioby

I don't mean to be picky,but you left out a square root sign...

Daniel.

6. Feb 21, 2005

### ehild

Oppps, really, you are right....

ehild