Is this a good substitution that will work

[rock.freak667]

Homework Statement

Prove \int_0^{1} \frac{1}{\sqrt{x^2+6x+25}} = ln(\frac{1+\sqrt{2}}{2})

Homework Equations

The Attempt at a Solution

\int_0^{1} \frac{1}{\sqrt{x^2+6x+25}}<br /> <br /> = \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

 

[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.

 

[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.

 

[HallsofIvy]

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,
\frac{x+3}{4}[/itex].