1. ## Integral Substitution

Made the problem nice and neat with Word Tools: I have a midterm on monday and I can SOLVE ANY problem and have memorized all common trig integrals. This is the only thing I can't do and its frustrating me and I forgot to ask my professor in office hours. Ive done every homework assignment twice, im intent on acing this class, and this is a harder problem on the review for the test.

http://img504.imageshack.us/img504/1460/calc2la2.png

I have tried all sorts of substitutions, even completeing the square and then using trig identities, Ive been working non stop on this problem for 40 minutes..

2. The imageshack link went down so I made a direct attachment.

3. Hello,

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}}$

Substitute $t=\sqrt{x^2-2x-8}$
$\implies t^2=x^2-2x-8 \implies 2t dt=2(x-1) dx \implies dx=\frac{t}{x-1} dt$

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=\int \frac xt \times \frac{t}{x-1} ~ dt$

$=\int \frac{x}{x-1} ~ dt=\int 1+\frac{1}{x-1} ~dt=t+\int \frac{1}{x-1} ~dt$

In order to calculate the last integral, you must have x-1 as a function of t.

$t^2=x^2-2x-8=(x-1)^2-9 \implies t^2+9=(x-1)^2 \implies x-1=\sqrt{t^2+9}$

So $\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt=t+\int \frac{1/3}{\sqrt{(t/3)^2+1}} ~dt$

Recall $\int \frac{1}{\sqrt{1+x^2}}=\text{arcsinh}(x)+\mathcal{ C}$

And after that, substitute back t=sqrt(x^2-2x-8)

4. Originally Posted by Moo
Hello,

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}}$

Substitute $t=\sqrt{x^2-2x-8}$
$\implies t^2=x^2-2x-8 \implies 2t dt=2(x-1) dx \implies dx=\frac{t}{x-1} dt$

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=\int \frac xt \times \frac{t}{x-1} ~ dt$

$=\int \frac{x}{x-1} ~ dt=\int 1+\frac{1}{x-1} ~dt=t+\int \frac{1}{x-1} ~dt$

In order to calculate the last integral, you must have x-1 as a function of t.

$t^2=x^2-2x-8=(x-1)^2-9 \implies t^2+9=(x-1)^2 \implies x-1=\sqrt{t^2+9}$

So $\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt=t+\int \frac{1/3}{\sqrt{({\color{red}t}/3)^2+1}} ~dt$

Recall $\int \frac{{\color{red}dt}}{\sqrt{1+{\color{red}t}^2}}= \text{arcsinh}({\color{red}t})+\mathcal{C}$

And after that, substitute back t=sqrt(x^2-2x-8)
A couple of small copy and paste errors corrected in red.

5. Originally Posted by mr fantastic
A couple of small copy and paste errors corrected in red.
Thanks
Just the first one actually =)

6. Originally Posted by Moo
Hello,

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}}$

[snip
Another method:

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}} = \int \frac{(x - 1) + 1}{\sqrt{x^2-2x-8}} \, dx = \int \frac{x - 1}{\sqrt{x^2-2x-8}} \, dx + \int \frac{1~dx}{\sqrt{x^2-2x-8}}$.

The first integral is simple: Make the substitution $u = x^2 - 2x - 8$

The second integral is simple if you're familiar with hyperbolic functions but not so simple if you're not:

$\int \frac{1~dx}{\sqrt{x^2-2x-8}} = \int \frac{1~dx}{\sqrt{(x - 1)^2 - 3^2}} = \frac{1}{3} \int \frac{1~dx}{\sqrt{\left(\frac{x - 1}{3}\right)^2 - 1}} = \frac{1}{3} \cosh^{-1} \left( \frac{x-1}{3}\right) + C$.

It can also be done using a trig substitution but that's a pain.

Note: $\cosh^{-1} A = \ln \left( A + \sqrt{A^2 - 1} \right)$ which would explain why the answer can also be expressed using logarithms:

7. hmm interesting, we are allowed to leave multiple variables within the integral as long as they cancel?

hmm.. thats not the answer it gives on the website, its expressed in "ln". We are not required to memorize the natural log formulas, but just by speculation I dont think it would equal his answer, but it looks right to me.

interesting I never thought you could have the "dt" and the actual variable differ, man.. all these tricks and stuff

thanks guys,

8. OK i found it out completely using trig subs,

I set $t = 3sec\theta\$ for only the integral

at this step

$
\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt
$

then derived such and so forth and got the answer.. wow that problem is tedious the first tiem on paper.. 1 page

OK i found it out completely using trig subs,

I set $t = 3sec\theta\$ for only the integral

at this step

$
\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt
$

then derived such and so forth and got the answer.. wow that problem is tedious the first tiem on paper.. 1 page
Originally Posted by mr fantastic
Another method:

$\int \frac{x ~dx}{\sqrt{x^2-2x-8}} = \int \frac{(x - 1) + 1}{\sqrt{x^2-2x-8}} \, dx = \int \frac{x - 1}{\sqrt{x^2-2x-8}} \, dx + \int \frac{1~dx}{\sqrt{x^2-2x-8}}$.

The first integral is simple: Make the substitution $u = x^2 - 2x - 8$

The second integral is simple if you're familiar with hyperbolic functions but not so simple if you're not:

$\int \frac{1~dx}{\sqrt{x^2-2x-8}} = \int \frac{1~dx}{\sqrt{(x - 1)^2 - 3^2}} = \frac{1}{3} \int \frac{1~dx}{\sqrt{\left(\frac{x - 1}{3}\right)^2 - 1}} = \frac{1}{3} \cosh^{-1} \left( \frac{x-1}{3}\right) + C$.

It can also be done using a trig substitution but that's a pain.

[snip]
Told you so!

10. sorry got clouded in the whole mess you guys posted trying to decipher it

you're guys method was pretty advanced, I dont think I have ever encountered a problem this hard yet, glad I came here for a fast response, well test tomorrow.. learning the reduction formulas for the 10 pts of extra credit, hoping for 130/120 lets hope

Also I prefer trig methods for some reason

Thanks again