1. Sep 30, 2004

nuub

I am having problem with two math questions that I have been asked to try and although I do not need to do them they are really frustrating me.

First off intergrate (x-1)/squareroot(4x^2-1)

I am sure this is something to do with replacing x to get cos^2 (x)-1 so that this becomes 2cos(2x) but I can't intergrate from there either.

Also intergrate 1/squareroot(1+8x-4x^2) again I am sure it is a substitution for trig but I just can't see it.

Any help would be greatly appreciated. Sorry I caan't lay the questions out any better

2. Sep 30, 2004

Zurtex

I'm sorry to say but for problems like this simple trig substations will not service, a bit of algebraic manipulation to make things look nicer is in order:

For the 1st one you should realise that:

$$\int \frac{x-1}{\sqrt{4x^2-1}} dx = \int \frac{x}{\sqrt{4x^2-1}} - \frac{1}{\sqrt{4x^2-1}}dx$$

Now if you split the integral up:

$$\int \frac{x}{\sqrt{4x^2-1}}dx - \int\frac{1}{\sqrt{4x^2-1}}dx$$

Not wanting to do the whole thing for you I'll leave you to work out the rest.

As for the second the trick I belive is completing the square:

$$\int \frac{1}{\sqrt{1+8x-4x^2}} dx= \int \frac{1}{\sqrt{5-4(x - 1)^2}}dx$$

Then use the substation u = x - 1 and again I'll leave you to do the rest. (I would seriously check my arithmetic if I were you, not feeling that well on it tonight).

3. Sep 30, 2004

nuub

thanks

thank you for the quick responce. I can not believe I missed I really hate it when I overlook the obvious.

Thanks