# Is this a good substitution that will work

1. Sep 13, 2007

### rock.freak667

1. The problem statement, all variables and given/known data
Prove $$\int_0^{1} \frac{1}{\sqrt{x^2+6x+25}} = ln(\frac{1+\sqrt{2}}{2})$$

2. Relevant equations

3. The attempt at a solution

$$\int_0^{1} \frac{1}{\sqrt{x^2+6x+25}} = \int_0^{1} \frac{1}{\sqrt{(x+3)^2+16}}$$

Let $$x+3=4tan\theta$$ so that $$dx=4sec^2\theta d\theta$$

and so the problem becomes

$$\int \frac{4sec^2\theta}{\sqrt{16tan^2\theta+16}} d\theta$$

giving $$\int sec\theta d\theta = ln|sec\theta + tan\theta|+ K$$

Last edited: Sep 13, 2007
2. Sep 14, 2007

### danago

Isolate theta in the substitution you made i.e. $$\theta=arctan\frac{x+3}{4}$$. From there, you should be able to evaluate the definite integral, and come to the required solution.

3. Sep 14, 2007

### Gib Z

You are correct so far, now just change the bounds accordingly to your substitutions. The first bound, x=1, so put that into your subsitution, tan theta = 1, ie theta = pi/4. Do the same for the other bound, and evaluate from your last line.

4. Sep 14, 2007

### HallsofIvy

Staff Emeritus
It is not absolutely necessary to let
$$\theta=arctan\frac{x+3}{4}$$
(and then use trig identities). Imagine a right triangle with one angle $\theta$ since you know
$$tan(\theta)= \frac{x+3}{4}$$,
the triangle has "opposite side" of length x+3 and "near side" of 4.

By the Pythagorean theorem, the square of the hypotenuse is $(x+3)^2+ 16= x^2+ 6x+ 25$.

Then $sec(\theta)$, hypotenuse over near side is $$\frac{x^2+ 6x+ 25}{4}$$
and
$$tan(\theta)$$
is, of course,
[tex]\frac{x+3}{4}[/itex].