Indefinite Integration Derivation ?

[f(x)]

Homework Statement

I want to prove the standard result for indefinite integrals that $$\int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--} \int {\frac{1}{a^2-x^2} .dx } \\$$
$$= \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\ = \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\$$
$$= \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\$$
$$= \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\$$

$$= \frac{1}{2a}\ln{(a-x)}+ln{(a+x)} .dx$$
I used this from an earlier derivation of $$\int{\frac{1}{x^2-a^2} } \mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} )$$
was converted to $$= \frac{1}{2a}\ln{(x-a)}-ln{(x-a)} .dx$$ similarly Plz suggest where i went wrong and suggest a way to get the desired proof
Thx

EDIT: There's some trouble with latex. The code i entered was -: (space to avoid repetition)
[i tex] \int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--}
\int {\frac{1}{a^2-x^2} .dx } \\
= \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\
= \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\
= \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\
= \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\
= \frac{1}{2a}\ln{a-x}+ln{a+x} .dx mbox{I used this from an earlier derivation of }
\int{\frac{1}{x^2-a^2} } \ mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} )
mbox{ was converted to ln similarly } [/i tex]
I am unable to get the complete msg?

 

[neutrino]

Have you tried trig. substitutions? If so, which one?

And, btw, asking others where you have gone wrong without showing your work will not work.

 

[f(x)]

Have you tried trig. substitutions? If so, which one?

And, btw, asking others where you have gone wrong without showing your work will not work.

I HAVE shown my work...its only that i can't get it working with latex. Anyways its more or less working..so please check again :)

 

[Dick]

If your final answer is (ln(a-x)+ln(a+x))/(2*a), there is a sign error on the first log. It should be negative. (Why?). Then if you combine the logs using ln(a)-ln(b)=ln(a/b) you get exactly what you are looking for.

 

[f(x)]

If your final answer is (ln(a-x)+ln(a+x))/(2*a), there is a sign error on the first log. It should be negative. (Why?). Then if you combine the logs using ln(a)-ln(b)=ln(a/b) you get exactly what you are looking for.

Ahh fine...i got it. Its s'posed to be negatibe because in 1/(a-x) , x is -ve
Thx for the help

 

[neutrino]

I HAVE shown my work...its only that i can't get it working with latex. Anyways its more or less working..so please check again :)

Oh, sorry about that. I guess Dick has provided you the answer.

 

[AngeloG]

As said: there is a trig substitution that you can use that will help out a lot =). Simply, draw a triangle and you will understand what you can substitute it with.

You should end up with a trig function, sin, cos, csc, sec, tan, cot (x). From there, you can go back to the original equation based on your choice heh.

 

[D H]

As said: there is a trig substitution that you can use that will help out a lot =). Simply, draw a triangle and you will understand what you can substitute it with.

If you had read the previous posts, you would have seen that this problem has been solved without trig substitutions. He simply multiplied the original integrand by one and added zero (rather creatively), and voila, out came the result. The only problem in the original post was in computing $\int \frac 1{a-x} dx = \ln(a-x)$.

 

[Dick]

If you actually do a trig substitution on this, you'll see that it is really the long way around.

 

[f(x)]

Yeah I thought about trig substitution...IMO i think it has to be either $$x=acos\Theta$$ or $$x=asin \Theta$$ But on solving further it still leaves $$sec\Theta or cosec\Theta$$ to be integrated, which gives something in form of $$\frac{1}{a} \ln{\mid cosec\Theta-cot\Theta \mid} +c$$

 

[dextercioby]

You can also use the change of variable $x=a\tanh t$. The result is immediate.

 

[f(x)]

You can also use the change of variable $x=a\tanh t$. The result is immediate.

I m yet unfamiliar with hyperbolic functions...:shy:

 

[Gib Z]

What is it with dexter and his hyperbolic substitutions :P

Could you do us a favour and show us how it works? I haven't worked with hyperbolics much either.

 

[dextercioby]

$$\int \frac{dx}{a^2 -x^2}\rightarrow \frac{1}{a}\int \frac{dt}{\cosh^2 t} \ \frac{1}{1-\tanh^2 t}=\frac{t}{a}+C\rightarrow \frac{1}{a} \mbox{argtanh}\frac{x}{a} +C$$

 

[Gib Z]

assuming that's an "arc" instead of "arg", i see it now. thanks

 

[dextercioby]

It's "arg", not "arc" for good reasons.

 

[Gib Z]

O right that is so huh? Argument :) Forgot about my imaginary stuff for a while :)

 

[dextercioby]

Why do you think it's "arc" for "arctangent", "arcsine",... ?

 

[Gib Z]

Arcument :)