Question Number 222121 by wewji12 last updated on 18/Jun/25
$$\int\:\mathrm{acos}\left(\frac{\mathrm{cos}\left(\varrho\right)}{\mathrm{1}+\mathrm{2cos}\left(\varrho\right)}\right)\:\mathrm{d}\varrho \\ $$
Answered by Nicholas666 last updated on 19/Jun/25
$$\:\:\:\:\int{cos}^{−\mathrm{1}} \:\left(\frac{{cos}\:{x}}{\mathrm{1}\:+\mathrm{2}\:{cos}\:{x}}\right)\:{dx}\:=\:\int\:\frac{{cos}\:{x}}{{cox}\left(\mathrm{1}+\mathrm{2}\:{cos}\:{x}\right)}\:{dx}\:=\int\:\frac{\mathrm{1}\:}{\mathrm{1}\:+\mathrm{2}\:{cos}\:{x}}\:{dx} \\ $$$$\:\:\:\mathrm{let};\:\:\:{t}\:=\:{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow\:{cos}\:{x}\:=\:\frac{\mathrm{1}\:−{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:\mathrm{and}\:{dx}=\frac{\mathrm{2}}{\mathrm{1}\:+{t}^{\mathrm{2}} }\:{dt}.\:\: \\ $$$$\:\:\:=\:\:\int\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\:\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}\:\centerdot\:\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\: \\ $$$$\:\:=\:\int\:\frac{\mathrm{1}}{\frac{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\:}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\centerdot\:\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$$\:\:=\:\int\:\frac{\mathrm{1}\:+{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\:\centerdot\:\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:=\:\int\:\frac{\mathrm{2}}{\mathrm{3}−{t}^{\mathrm{2}} }\:{dt}\:=\:\mathrm{2}\int\frac{\mathrm{1}}{\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} }\:{dt}\:=\:\mathrm{2}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{tanh}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\mathrm{3}}}\right)\:+\:{C} \\ $$$$\:\:\:\: \\ $$$$\int{cos}^{−\mathrm{1}} \left(\frac{{cos}\:{x}}{\mathrm{1}\:+\:\mathrm{2}\:{cos}\:{x}}\right)\:{dx}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tanh}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\:+\:{Constants} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 20/Jun/25
$$\mathrm{How}\:\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{2cos}\:{x}}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2cos}\:{x}}\:??? \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{wronger}\:\mathrm{than}\:\mathrm{wrong}… \\ $$
Commented by MathematicalUser2357 last updated on 20/Jun/25
$${Why}\:{is}\:\int\frac{\mathrm{cos}\:{x}}{\mathrm{cos}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)}{dx}=\int\frac{\mathrm{cos}\:{x}}{{cox}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)}{dx}? \\ $$
Commented by Nicholas666 last updated on 20/Jun/25
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{correcting}\:\mathrm{me}, \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{fatal}\:\mathrm{mistake}\:\mathrm{I}\:\mathrm{have}\:\mathrm{ever}\:\mathrm{made}, \\ $$$$\mathrm{But}\:\mathrm{this}\:\mathrm{is}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{no}\:\mathrm{elementary}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{in}\:\mathrm{closed}\:\mathrm{form}\:.\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Commented by Frix last updated on 20/Jun/25
$$\mathrm{The}\:\mathrm{result}\:\mathrm{is}\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{24}}\:\mathrm{but}\:\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{prove}\:\mathrm{until} \\ $$$$\mathrm{now}. \\ $$
Commented by Nicholas666 last updated on 20/Jun/25
$$\mathrm{the}\:\mathrm{ans}\:\mathrm{you}\:\mathrm{said}\:\mathrm{is}\:\mathrm{wrong},\:\mathrm{the}\:\mathrm{ans}\:\mathrm{you}\:\mathrm{gave}\:\mathrm{is}\:\mathrm{if}\:\mathrm{it}\:\mathrm{has}\:\mathrm{a}\:\mathrm{limit}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by Frix last updated on 20/Jun/25
![Yes. The same question has been posted several times before with the limits [0; (π/2)].](https://www.tinkutara.com/question/Q222229.png)
$$\mathrm{Yes}.\:\mathrm{The}\:\mathrm{same}\:\mathrm{question}\:\mathrm{has}\:\mathrm{been}\:\mathrm{posted} \\ $$$$\mathrm{several}\:\mathrm{times}\:\mathrm{before}\:\mathrm{with}\:\mathrm{the}\:\mathrm{limits}\:\left[\mathrm{0};\:\frac{\pi}{\mathrm{2}}\right]. \\ $$
Answered by MrGaster last updated on 27/Jun/25
![ϱ=θ ∫ acos(((cos(θ))/(1+2cos(θ)))) dθ =(π/2)∫_0 ^(π/2) dθ−Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1)))∫_0 ^(π/2) (((cos θ)/(1+2 cos θ)))^(2n+1) dθ =(π/2)∙(π/2)−Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1)))∙2∫_0 ^1 (((1−t^2 )^(2n+1) )/((3−t^2 )^(2n+1) (1+t^2 )))dt t=tan((θ/2)) cos θ=((1−t^2 )/(1+t^2 )),dθ=((2dt)/(1+t^2 )) θ:0 (π/2)⇒t:0→1 ((cos θ)/(1+2 cos θ))=(((1−t^2 )/(1+t^2 ))/(1+2∙((1−t^2 )/(1+t^2 ))))=((1−t^2 )/((1+t^2 )(1−t^2 )))=((1−t^2 )/(3−t^2 )) (((cos θ)/(1+2 cos θ)))^(2n+1) dθ=(((1−t^2 )/(3−t^2 )))^(2n+1) ((2dt)/(1+t^2 )) ∫_0 ^(π/2) (((cos θ)/(1+2 cos θ)))^(2n+1) dθ=∫_0 ^1 (((1−t^2 )/(3−t^2 )))^(2n+1) ((2dt)/(1+t^2 ))=2∫_0 ^1 (((1−t^2 )^(2n+1) )/((3−t^2 )^(2n+1) (1+t^2 )))dt Σ_(n=0) ^( ∞) ( (((2n)),(n) )/(4^n (2n+1)))∙2∫_0 ^1 (((1−t^2 )^(2n+1) )/((3−t^2 )^(2n+1) (1+t^2 )))dt=2∫_0 ^1 (1/(1+t^2 ))Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1)))(((1−t^2 )/(3−t^2 )))^(2n+1) dt arccos y=(π/2)−arcsin y,arcsin y=Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1))) fot ∣y∣<1 y=((1−t^2 )/(3−t^2 )),t∈[0,1]⇒y∈[0,(1/3)]⊂[−1,1] arcsin(((1−t^2 )/(3+t^2 )))=Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1)))(((1−t^2 )/(3−t^2 )))^(2n+1) Σ_(n=0) ^∞ ( (((2n)),(n) )/(4^n (2n+1)))(((1−t^2 )/(3−t^2 )))^(2n+1) =arcsin(((1−t^2 )/(3−t^2 ))) 2∫_0 ^1 (1/(1+t^2 ))arcsin(((1−t^2 )/(3−t^2 )))dt arcsin(((1−t^2 )/(3−t^2 )))=arcsin(((−(t^2 +1))/(−(t^2 +3))))=arcsin(((t^2 −1)/(t^2 −3))) ((t^2 −1)/(t^2 −3))=1+(2/(t^3 −3)) arcsin(1+(2/(t^2 −3))) u=t^2 ,du=2tdt,dt=(du/(2(√u))) t:0→1⇒u:0→1 2∫_0 ^1 (1/(1+u))arcsin(((u−1)/(u−3)))(du/(2(√u)))=∫_0 ^1 (1/((1+u)(√u)))arcsin(((u−1)/(u−3)))du ((u−1)/(u−3))=1+(2/(u−1)) u=(√u),u=v^2 ,du=2vdv u:0→1 ∫(1/((1+v^2 )v))arcsin(((v^2 −1)/(v^2 −3)))2dv=2∫_0 ^1 (1/(1−v^2 ))arcsin(((v^2 −1)/(v^2 −3)))dv ((v^2 −1)/(v^2 −3))=((−(1−v^2 ))/(−(3−v^2 )))=((1−v^2 )/(3−v^2 )) arcsin(((1−v^2 )/(3−v^2 ))) φ=arcsin v,v=sin φ,dv=cos φ dφ,φ:0→(π/2) 2∫_0 ^(π/2) (1/(1+sin^2 φ))arcsin(((1−sin^2 φ)/(3−sin^2 φ)))cos φdφ 1−sin^2 φ=cos^2 φ,3−sin^2 φ=2+cos^2 φ arcsin(((cos^2 φ)/(2+cos^2 φ))) 2∫_0 ^(π/2) ((cos φ)/(1+sin^2 φ))arcsin(((cos^2 φ)/(2+cos^2 φ)))dφ w=sin φ,dv=cos φdφ,φ:0→(π/2)⇒w:0→1 2∫_0 ^1 (1/(1+w^2 ))arcsin(((1−w^2 )/(2+1−w^2 )))dw=2∫_0 ^1 (1/(1+w^2 ))arcsin(((1−w^2 )/(3−w^2 )))dw arcsin(((1−w^2 )/(3−w^2 )))=arctan(((√((3−w^2 )^2 −(1−w^2 )^2 ))/(1−w^2 ))) (3−w^2 )−(1−w^2 )=(3−w^2 −1+w^2 )(3−w^2 +1−w^2 )=(2)(4−2w^2 )=8−4w^2 (√(8−4w^2 ))=2(√2)(√(1−(w^2 /2))) arcsin(((1−w^2 )/(3−w^2 )))=arcsin(((2(√2)(√(1−(w^2 /(2+)))))/(1−w^2 ))) 1−w^2 =(1−w)(1+w) arctan((((√2)(√(2−w^2 )))/((1−w)(1+w)))) ∫_0 ^1 (1/(1+w^2 ))arctan((((√2)(√(2−w^2 )))/((1−w)(1+w))))dw ζ(2)=(π^2 /6) ∫_0 ^(π/2) (1/(1+w^2 ))arctan(((cos θ)/(1+2 cos θ)))dθ=(π^2 /4)−2∫_0 ^1 (1/(1+w^2 ))arcsin(((1−w^2 )/(3−w^2 )))dw =(π^2 /4)−2∙(π^2 /(48)) =((12π^2 )/(48))−((2π^2 )/(48))=((10π^2 )/(48))=((5π^2 )/(24))](https://www.tinkutara.com/question/Q222456.png)
$$\varrho=\theta \\ $$$$\int\:\mathrm{acos}\left(\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{1}+\mathrm{2cos}\left(\theta\right)}\right)\:\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {d}\theta−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}{n}+\mathrm{1}} {d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{2}}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\centerdot\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{3}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$${t}=\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} },{d}\theta=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\theta:\mathrm{0} \frac{\pi}{\mathrm{2}}\Rightarrow{t}:\mathrm{0}\rightarrow\mathrm{1} \\ $$$$\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta}=\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{2}\centerdot\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}{n}+\mathrm{1}} {d}\theta=\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}{n}+\mathrm{1}} \frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}{n}+\mathrm{1}} {d}\theta=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}{n}+\mathrm{1}} \frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{3}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\:\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\centerdot\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{3}−{t}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}{n}+\mathrm{1}} {dt} \\ $$$$\mathrm{arccos}\:{y}=\frac{\pi}{\mathrm{2}}−\mathrm{arcsin}\:{y},\mathrm{arcsin}\:{y}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\:\mathrm{fot}\:\mid{y}\mid<\mathrm{1} \\ $$$${y}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} },{t}\in\left[\mathrm{0},\mathrm{1}\right]\Rightarrow{y}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{3}}\right]\subset\left[−\mathrm{1},\mathrm{1}\right] \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}+{t}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}{n}+\mathrm{1}} =\mathrm{arcsin}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\mathrm{arcsin}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right){dt} \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}−{t}^{\mathrm{2}} }\right)=\mathrm{arcsin}\left(\frac{−\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{−\left({t}^{\mathrm{2}} +\mathrm{3}\right)}\right)=\mathrm{arcsin}\left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{3}}\right) \\ $$$$\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{3}}=\mathrm{1}+\frac{\mathrm{2}}{{t}^{\mathrm{3}} −\mathrm{3}} \\ $$$$\mathrm{arcsin}\left(\mathrm{1}+\frac{\mathrm{2}}{{t}^{\mathrm{2}} −\mathrm{3}}\right) \\ $$$${u}={t}^{\mathrm{2}} ,{du}=\mathrm{2}{tdt},{dt}=\frac{{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$${t}:\mathrm{0}\rightarrow\mathrm{1}\Rightarrow{u}:\mathrm{0}\rightarrow\mathrm{1} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{u}}\mathrm{arcsin}\left(\frac{{u}−\mathrm{1}}{{u}−\mathrm{3}}\right)\frac{{du}}{\mathrm{2}\sqrt{{u}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{u}\right)\sqrt{{u}}}\mathrm{arcsin}\left(\frac{{u}−\mathrm{1}}{{u}−\mathrm{3}}\right){du} \\ $$$$\frac{{u}−\mathrm{1}}{{u}−\mathrm{3}}=\mathrm{1}+\frac{\mathrm{2}}{{u}−\mathrm{1}} \\ $$$${u}=\sqrt{{u}},{u}={v}^{\mathrm{2}} ,{du}=\mathrm{2}{vdv} \\ $$$${u}:\mathrm{0}\rightarrow\mathrm{1} \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}+{v}^{\mathrm{2}} \right){v}}\mathrm{arcsin}\left(\frac{{v}^{\mathrm{2}} −\mathrm{1}}{{v}^{\mathrm{2}} −\mathrm{3}}\right)\mathrm{2}{dv}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{v}^{\mathrm{2}} }\mathrm{arcsin}\left(\frac{{v}^{\mathrm{2}} −\mathrm{1}}{{v}^{\mathrm{2}} −\mathrm{3}}\right){dv} \\ $$$$\frac{{v}^{\mathrm{2}} −\mathrm{1}}{{v}^{\mathrm{2}} −\mathrm{3}}=\frac{−\left(\mathrm{1}−{v}^{\mathrm{2}} \right)}{−\left(\mathrm{3}−{v}^{\mathrm{2}} \right)}=\frac{\mathrm{1}−{v}^{\mathrm{2}} }{\mathrm{3}−{v}^{\mathrm{2}} } \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{1}−{v}^{\mathrm{2}} }{\mathrm{3}−{v}^{\mathrm{2}} }\right) \\ $$$$\phi=\mathrm{arcsin}\:{v},{v}=\mathrm{sin}\:\phi,{dv}=\mathrm{cos}\:\phi\:\mathrm{d}\phi,\phi:\mathrm{0}\rightarrow\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \phi}\mathrm{arcsin}\left(\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \phi}{\mathrm{3}−\mathrm{sin}^{\mathrm{2}} \phi}\right)\mathrm{cos}\:\phi{d}\phi \\ $$$$\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \phi=\mathrm{cos}^{\mathrm{2}} \phi,\mathrm{3}−\mathrm{sin}^{\mathrm{2}} \phi=\mathrm{2}+\mathrm{cos}^{\mathrm{2}} \phi \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{cos}^{\mathrm{2}} \phi}{\mathrm{2}+\mathrm{cos}^{\mathrm{2}} \phi}\right) \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:\phi}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \phi}\mathrm{arcsin}\left(\frac{\mathrm{cos}^{\mathrm{2}} \phi}{\mathrm{2}+\mathrm{cos}^{\mathrm{2}} \phi}\right){d}\phi \\ $$$${w}=\mathrm{sin}\:\phi,{dv}=\mathrm{cos}\:\phi{d}\phi,\phi:\mathrm{0}\rightarrow\frac{\pi}{\mathrm{2}}\Rightarrow{w}:\mathrm{0}\rightarrow\mathrm{1} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{w}^{\mathrm{2}} }\mathrm{arcsin}\left(\frac{\mathrm{1}−{w}^{\mathrm{2}} }{\mathrm{2}+\mathrm{1}−{w}^{\mathrm{2}} }\right){dw}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{w}^{\mathrm{2}} }\mathrm{arcsin}\left(\frac{\mathrm{1}−{w}^{\mathrm{2}} }{\mathrm{3}−{w}^{\mathrm{2}} }\right){dw} \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{1}−{w}^{\mathrm{2}} }{\mathrm{3}−{w}^{\mathrm{2}} }\right)=\mathrm{arctan}\left(\frac{\sqrt{\left(\mathrm{3}−{w}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{1}−{w}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{1}−{w}^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{3}−{w}^{\mathrm{2}} \right)−\left(\mathrm{1}−{w}^{\mathrm{2}} \right)=\left(\mathrm{3}−{w}^{\mathrm{2}} −\mathrm{1}+{w}^{\mathrm{2}} \right)\left(\mathrm{3}−{w}^{\mathrm{2}} +\mathrm{1}−{w}^{\mathrm{2}} \right)=\left(\mathrm{2}\right)\left(\mathrm{4}−\mathrm{2}{w}^{\mathrm{2}} \right)=\mathrm{8}−\mathrm{4}{w}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{8}−\mathrm{4}{w}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−\frac{{w}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\mathrm{arcsin}\left(\frac{\mathrm{1}−{w}^{\mathrm{2}} }{\mathrm{3}−{w}^{\mathrm{2}} }\right)=\mathrm{arcsin}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−\frac{{w}^{\mathrm{2}} }{\mathrm{2}+}}}{\mathrm{1}−{w}^{\mathrm{2}} }\right) \\ $$$$\mathrm{1}−{w}^{\mathrm{2}} =\left(\mathrm{1}−{w}\right)\left(\mathrm{1}+{w}\right) \\ $$$$\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}\sqrt{\mathrm{2}−{w}^{\mathrm{2}} }}{\left(\mathrm{1}−{w}\right)\left(\mathrm{1}+{w}\right)}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{w}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}\sqrt{\mathrm{2}−{w}^{\mathrm{2}} }}{\left(\mathrm{1}−{w}\right)\left(\mathrm{1}+{w}\right)}\right){dw} \\ $$$$\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+{w}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta}\right){d}\theta=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{w}^{\mathrm{2}} }\mathrm{arcsin}\left(\frac{\mathrm{1}−{w}^{\mathrm{2}} }{\mathrm{3}−{w}^{\mathrm{2}} }\right){dw} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\centerdot\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$=\frac{\mathrm{12}\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{48}}=\frac{\mathrm{10}\pi^{\mathrm{2}} }{\mathrm{48}}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$