intergrate (2x-3)/((9-x^2)^1/2) dx
Believe I need to break this down into 2x/(9-x^2)^1/2dx - 3/(9-x^2)^1/2 dx. I cant afford to not undertstand these problems though.
At first I saw this problem and that the denominator could be written as the difference of two squares and then split by partial fractions which gives two integrals who are of the differential of a function over a function and are therefore logarithms...however this is rubbish as far as you are concerned because of that sneaky little square root I failed to note first time round!!
What you need is a sneaky little substitution, when you have integrals similar to this all you need to know is the trig identity $\displaystyle cos^2+sin^2=1$ (and the similar cosh identity is useful for others with - instead of +). You should notice that with this substitution (try x proportional to cos) you can get the whole root to cancel with the other contribution (remember if you sub x=f(y) then you must replace dx=f'(y)dy) which gives a nice easy integral...
If you have any problems with this lot let me know and I'll give you a few more clues (or if you want to check your answer i've got it on my scratch pad next to me) but I don't want to spoil your fun of learning this stuff. Besides, its fun to play the sadist who used to make me work things out for myself!!!
Hello, Bust20001
Intergrate: .$\displaystyle \int\frac{2x-3}{9-x^2)^{\frac{1}{2}}}\,dx$
Believe I need to break this down into: .$\displaystyle \int\frac{2x}{(9-x^2)^{\frac{1}{2}}}\,dx-\int\frac{3}{(9-x^2)^{\frac{1}{2}}}\,dx$ . Right!
The first integral is: .$\displaystyle \int(9-x^2)^{-\frac{1}{2}}dx$
. . Let $\displaystyle u\:=\:9-x^2\quad\Rightarrow\quad du\:=\:-2x\,dx\quad\hdots\;\;\text{etc.}$
The second integral is:.$\displaystyle \int\frac{dx}{\sqrt{9-x^2}}\quad\hdots\quad\text{arcsine}$