1. ## Substitution

Good morning,

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

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

I am coming up with:

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

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

I believe this should give me:

$\int{ududdu}$

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

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

Then take the derivative - don't forget to apply the chain rule:
$
\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}}
$

$
du = \frac {-dx} {cos^{-1} x\sqrt {1-x^2}}
$

So now you have:

$
\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 $u = cos^{-1} x$:

$
u=cos^{-1}x
$

$
\frac {du} {dx} = \frac {d(cos^{-1} x)} {dx} = \frac {-1} {\sqrt {1-x^2}}
$

$
du = \frac {-1} {\sqrt {1-x^2}}dx
$

$
\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!