1. ## Integration (2)

Using the substitution x = 1/y or otherwise, find ∫ {1/[x (x^2 - 1)^0.5]} dx.

Thank you!!

2. Hello
Originally Posted by Tangera
Using the substitution x = 1/y or otherwise, find ∫ {1/[x (x^2 - 1)^0.5]} dx.
Substituting $x=1/y$ you should get $-\int \frac{\mathrm{d}y}{\sqrt{1-y^2}}$. Now recall that the derivative of $y\mapsto \arccos y$ is $y\mapsto \frac{-1}{\sqrt{1-y^2}}$ and you are done.

3. $\int \frac{1}{x\sqrt{x^{2}-1}} \: dx$

$x = \frac{1}{y} \: \: \Rightarrow \: \: dx = -\frac{1}{y^2} \: dy$

Making the sub:
$\int \frac{1}{\frac{1}{y}\sqrt{\left(\frac{1}{y}\right) ^{2} - 1}} \: \left(-\frac{1}{y^2} \: dy\right) \: \: = \: \: -\int \frac{1}{y\sqrt{\frac{1}{y^2} - 1}} \: dy$

Now, imagine: $y = \sqrt{y^2}$

$-\int \frac{1}{\sqrt{y^{2}}\sqrt{\frac{1}{y^2} - 1}} \: dy \: \: = \: \: -\int \frac{1}{\sqrt{1 - y^{2}}} \: dy$

Does arccos/arcsin ring a bell?

4. $\arccos \frac{1}{x}$ only works for positive x. Remember, $\sqrt{x^2} = |x| \color{red}\neq x$

I have a more general solution.

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

Now I'll do something that you'll understand the reason later.

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

Let $u=\sqrt{x^2-1}$. Then $u^2 = x^2-1$. And $u~du = x~dx$. Now substitute everything,

$\int \frac{u}{(u^2+1)u}~du = \int \frac{1}{u^2+1}~du = \arctan u+C$

$= \arctan \sqrt{x^2-1}+C$