How do you evaluate int arccosx/sqrt(1-x^2) from [0, 1/sqrt2]?

Jan 18, 2017

${\int}_{0}^{\frac{1}{\sqrt{2}}} \arccos \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \frac{3 {\pi}^{2}}{32}$

Explanation:

$I = {\int}_{0}^{\frac{1}{\sqrt{2}}} \arccos \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

It's important to know that $\frac{d}{\mathrm{dx}} \arccos x = \frac{- 1}{\sqrt{1 - {x}^{2}}}$. So, if we let $u = \arccos x$ then $\mathrm{du} = \frac{- 1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$.

Also note that the bounds will change:

$x = \frac{1}{\sqrt{2}} \text{ "=>" } u = \arccos \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$

$x = 0 \text{ "=>" } u = \arccos \left(0\right) = \frac{\pi}{2}$

Then:

$I = - {\int}_{0}^{\frac{1}{\sqrt{2}}} \arccos x \left(\frac{- 1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}\right)$

$I = - {\int}_{\frac{\pi}{2}}^{\frac{\pi}{4}} u \textcolor{w h i t e}{.} \mathrm{du}$

$I = {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} u \textcolor{w h i t e}{.} \mathrm{du}$

$I = \frac{1}{2} {u}^{2} {|}_{\pi / 4}^{\pi / 2}$

$I = \frac{1}{2} \left({\left(\frac{\pi}{2}\right)}^{2} - {\left(\frac{\pi}{4}\right)}^{2}\right)$

$I = \frac{1}{2} \left({\pi}^{2} / 4 - {\pi}^{2} / 16\right)$

$I = \frac{1}{2} \left(\frac{4 {\pi}^{2} - {\pi}^{2}}{16}\right)$

$I = \frac{3 {\pi}^{2}}{32}$