1. ## Substitution

Good morning,

I am looking at the following function, trying to understand what value I should be substituting $\displaystyle u$ for.

$\displaystyle \int\frac{{\ln}(cos^{-1}x)dx}{cos^{-1}x\sqrt{1-x^2}}$

I am coming up with:

$\displaystyle u=\ln(cos^{-1}x)$

$\displaystyle du=\frac{1}{cos^{-1}x}$

I believe this should give me:

$\displaystyle \int{ududdu}$

More importantly, I am really struggling with the "why" part of this substitution. I must be missing something obvious.

2. Not quite right. If you set

$\displaystyle u=ln(cos^{-1}x)$

Then take the derivative - don't forget to apply the chain rule:
$\displaystyle \frac {du} {dx} = \frac 1 {cos^{-1} x} \cdot \frac {d(cos^{-1} x)} {dx} = \frac 1 {cos^{-1} x} \cdot \frac {-1} {\sqrt {1-x^2}}$
$\displaystyle du = \frac {-dx} {cos^{-1} x\sqrt {1-x^2}}$

So now you have:

$\displaystyle \int \frac {ln(cos^{-1}x ) dx} {cos^{-1}x \sqrt {1 - x^2}} = -\int u du =-\frac 1 2 u^2 + C = -\frac 1 2 (ln(cos^{-1}x)^2 + C$

You could have also done this by setting $\displaystyle u = cos^{-1} x$:

$\displaystyle u=cos^{-1}x$
$\displaystyle \frac {du} {dx} = \frac {d(cos^{-1} x)} {dx} = \frac {-1} {\sqrt {1-x^2}}$
$\displaystyle du = \frac {-1} {\sqrt {1-x^2}}dx$
$\displaystyle \int \frac {ln(cos^{-1}x ) dx} {cos^{-1}x \sqrt {1 - x^2}} = - \int \frac {ln(u)} u du = - \frac 1 2 (ln(u))^2 + C = - \frac 1 2 (ln(cos^{-1}x))^2 + C$

Always happy to help out a fellow ME!