How do you evaluate $\int \frac{arccos x}{\sqrt{1 - x^{2}}}$ $int arccosx/sqrt(1-x^2)$ from $[0, \frac{1}{\sqrt{2}}]$ $[0, 1/sqrt2]$ ?

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How do you evaluate $\int \frac{arccos x}{\sqrt{1 - x^{2}}}$ $int arccosx/sqrt(1-x^2)$ from $[0, \frac{1}{\sqrt{2}}]$ $[0, 1/sqrt2]$ ?

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Answer 1 · 2017-01-18T21:21:17

$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{arccos x}{\sqrt{1 - x^{2}}} d x = \frac{3 π^{2}}{32}$ $int_0^(1/sqrt2)arccosx/sqrt(1-x^2)dx=(3pi^2)/32$

Explanation:

$I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{arccos x}{\sqrt{1 - x^{2}}} d x$ $I=int_0^(1/sqrt2)arccosx/sqrt(1-x^2)dx$

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

Also note that the bounds will change:

$x = \frac{1}{\sqrt{2}} \Rightarrow u = arccos (\frac{1}{\sqrt{2}}) = \frac{π}{4}$ $x=1/sqrt2" "=>" "u=arccos(1/sqrt2)=pi/4$

$x = 0 \Rightarrow u = arccos (0) = \frac{π}{2}$ $x=0" "=>" "u=arccos(0)=pi/2$

Then:

$I = - \int_{0}^{\frac{1}{\sqrt{2}}} arccos x (\frac{- 1}{\sqrt{1 - x^{2}}} d x)$ $I=-int_0^(1/sqrt2)arccosx((-1)/sqrt(1-x^2)dx)$

$I = - \int_{\frac{π}{2}}^{\frac{π}{4}} u . d u$ $I=-int_(pi/2)^(pi/4)u color(white).du$

$I = \int_{\frac{π}{4}}^{\frac{π}{2}} u . d u$ $I=int_(pi/4)^(pi/2)ucolor(white).du$

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

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

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

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

$I = \frac{3 π^{2}}{32}$ $I=(3pi^2)/32$

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How do you evaluate $\int \frac{arccos x}{\sqrt{1 - x^{2}}}$ $int arccosx/sqrt(1-x^2)$ from $[0, \frac{1}{\sqrt{2}}]$ $[0, 1/sqrt2]$ ?

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How do you evaluate $\int \frac{arccos x}{\sqrt{1 - x^{2}}}$ $int arccosx/sqrt(1-x^2)$ from $[0, \frac{1}{\sqrt{2}}]$ $[0, 1/sqrt2]$ ?