Question Number 218781 by Ghisom last updated on 15/Apr/25
$$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$
Answered by breniam last updated on 20/Apr/25
![∫_0 ^(π/4) arccos((√(2/(3−tan^2 x))))dx= ∫_0 ^(π/4) arctan((√((1−(2/(3−tan^2 x)))/(2/(3−tan^2 x)))))= ∫_0 ^(π/4) arctan((√((1−tan^2 x)/2)))dx= ∫_0 ^(1/( (√2))) ∫_0 ^(π/4) (1/(y^2 (1−tan^2 x)+1))×(√(1−tan^2 x))dxdy={x^− =tanx} ∫_0 ^(1/( (√2))) ∫_0 ^1 (1/(y^2 (1−x^2 )+1))×(√(1−x^2 ))×(1/(x^2 +1))dxdy={x^− =arcsin x}= ∫_0 ^(1/( (√2))) ∫_0 ^(π/2) (1/(y^2 cos^2 x+1))×cos x×(1/(sin^2 x+1))×cosxdxdy =∫_0 ^(1/( (√2))) ∫_0 ^(π/2) ((cos^2 x)/((y^2 cos^2 x+1)(sin^2 x+1)))dxdy={x^− =tan x} =∫_0 ^(1/( (√2))) ∫_0 ^∞ ((1/(x^2 +1))/(((y^2 /(x^2 +1))+1)((x^2 /(x^2 +1))+1)))×(1/(x^2 +1))dxdy= ∫_0 ^(1/( (√2))) ∫_0 ^∞ (1/((x^2 +y^2 +1)(2x^2 +1)))dxdy= ∫_0 ^(1/( (√2))) (1/(2y^2 +1))∫_0 ^∞ [(2/(2x^2 +1))−(1/(x^2 +y^2 +1))]dxdy= (π/2)∫_0 ^(1/( (√2))) (1/(2y^2 +1))[(√2)−(1/( (√(y^2 +1))))]dy= (π/2)arctan(1/( (√2)))−(π/2)∫_0 ^(1/( (√2))) (1/( (√(y^2 +1))(2y^2 +1)))dy={y^− =arctan y}= (π/2)arctan(1/( (√2)))−(π/2)∫_0 ^(arctan (1/( (√2)))) ((1/(cos y))/(2tan^2 y+1))dy= (π/2)arctan (1/( (√2)))−(π/2)∫_0 ^(arctan (1/( (√2)))) ((cos y)/(2sin^2 y+cos^2 y))dy= (π/2)arctan (1/( (√2)))−(π/2)∫_0 ^(arctan (1/( (√2)))) ((cos y)/(sin^2 y+1))dy={y^− =sin y}= (π/2)arctan (1/( (√2)))−(π/2)∫_0 ^((1/( (√2)))/( (√(3/2)))) (1/(y^2 +1))dy= (π/2)[arctan (1/( (√2)))−arctan (1/( (√3)))]= (π/2)[arctan ((((1/( (√2)))−(1/( (√3))))/(1+(1/( (√6))))))]=(π^2 /(24))](https://www.tinkutara.com/question/Q219228.png)
$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{arccos}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} {x}}}\right)\mathrm{d}{x}= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{arctan}\left(\sqrt{\frac{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} {x}}}{\frac{\mathrm{2}}{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} {x}}}}\right)= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{arctan}\left(\sqrt{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{2}}}\right)\mathrm{d}{x}= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)+\mathrm{1}}×\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}\mathrm{d}{x}\mathrm{d}{y}=\left\{\overset{−} {{x}}=\mathrm{tan}{x}\right\} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+\mathrm{1}}×\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }×\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}=\left\{\overset{−} {{x}}=\mathrm{arcsin}\:{x}\right\}= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+\mathrm{1}}×\mathrm{cos}\:{x}×\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{1}}×\mathrm{cos}{x}\mathrm{d}{x}\mathrm{d}{y} \\ $$$$=\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{cos}^{\mathrm{2}} {x}}{\left({y}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+\mathrm{1}\right)\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{1}\right)}\mathrm{d}{x}\mathrm{d}{y}=\left\{\overset{−} {{x}}=\mathrm{tan}\:{x}\right\} \\ $$$$=\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}{\left(\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)}×\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\frac{\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\right]\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}\left[\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}\right]\mathrm{d}{y}= \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{arctan}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\mathrm{1}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{y}=\left\{\overset{−} {{y}}=\mathrm{arctan}\:{y}\right\}= \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{arctan}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\frac{\mathrm{1}}{\mathrm{cos}\:{y}}}{\mathrm{2tan}^{\mathrm{2}} {y}+\mathrm{1}}\mathrm{d}{y}= \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\mathrm{cos}\:{y}}{\mathrm{2sin}^{\mathrm{2}} {y}+\mathrm{cos}^{\mathrm{2}} {y}}\mathrm{d}{y}= \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{\mathrm{cos}\:{y}}{\mathrm{sin}^{\mathrm{2}} {y}+\mathrm{1}}\mathrm{d}{y}=\left\{\overset{−} {{y}}=\mathrm{sin}\:{y}\right\}= \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}}} {\int}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}= \\ $$$$\frac{\pi}{\mathrm{2}}\left[\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{arctan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]= \\ $$$$\frac{\pi}{\mathrm{2}}\left[\mathrm{arctan}\:\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}}\right)\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$ \\ $$