# Math Help - How to simplify this?

1. ## How to simplify this?

In my book I have a proof for this integral: $\int\frac{dx}{\sqrt{a^2+x^2}}$

$\int\frac{dx}{\sqrt{a^2+x^2}}$ = $\int\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}}$ = $ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})$

${\frac {d}{dx}}ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})$ = $\frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{ 1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}$

How can I simplify = $\frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{ 1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}$ into $\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}}$ ?

2. You could actually perform the substitution, rather than carrying it in the notation where it seems to be confusing you.

Do you really mean "proof"? Maybe just a demonstration? The last marvelous step to the natural logarithm is a bit magic.

3. Originally Posted by Bart
In my book I have a proof for this integral: $\int\frac{dx}{\sqrt{a^2+x^2}}$

$\int\frac{dx}{\sqrt{a^2+x^2}}$ = $\int\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}}$ = $ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})$

${\frac {d}{dx}}ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})$ = $\frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{ 1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}$

How can I simplify = $\frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{ 1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}$ into $\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}}$ ?
That sucks!!! the way it's carried out there.

This is a prime candidate for tan subsitution.

$\int\frac{dx}{\sqrt{a^2+x^2}}$

Let $x = a tan \theta$ and $dx = a sec^2 \theta d \theta$

$\int\frac{dx}{\sqrt{a^2+x^2}} = \int \frac{ a sec^2 \theta }{ a sec \theta } d \theta = \int sec \theta d \theta = ln| sec \theta + tan \theta | + C$

We arrive at

$ln | \frac{ \sqrt{a^2 + x^2 } }{a} + \frac{x}{a} | + C$

And simplify to

$ln( \sqrt{a^2 + x^2} + x) + C_2$

Noting that $\sqrt{a^2 + x^2} + x > 0$ for all x, so we can disregard the absolute value

4. I don't totally hate what it was trying to do. With a little more experience in the sightly unusual notation, it isn't quite terrible.