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Question-119930

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Question Number 119930 by Lordose last updated on 28/Oct/20
Answered by mathmax by abdo last updated on 28/Oct/20
A =∫_0 ^1 arcsin(tanθ)dθ we do the changement arcsin(tanθ)=x ⇒tanθ =sinx ⇒θ =artan(sinx) ⇒ A =∫_0 ^(arcsin(tan1)) x ×((cosx)/(1+sin^2 x))dx ⇒ A =_(tan((x/2))=t) ∫_0 ^(tan(((arcsin(tan1))/2))) ((2arctant×((1−t^2 )/(1+t^2 )))/(1+((4t^2 )/((1+t^2 )^2 ))))×((2dt)/(1+t^2 )) =4∫_0 ^c (((1−t^2 )arctant)/((1+t^2 )^2 (1+((4t^2 )/((1+t^2 )^2 )))))dt =4∫_0 ^c (((1−t^2 )arctant)/((1+t^2 )^2 +4t^2 ))dt =4∫_0 ^c (((1−t^2 )arctan(t))/(t^4 +2t^2 +1+4t^2 ))dt =4∫_0 ^c (((1−t^2 )arctan(t))/(t^4 +6t^2 +1))dt let ϕ(a) =∫_0 ^c (((1−t^2 )arctan(at))/(t^4 +6t^2 +1))dt (c=tan(((arcsin(tan1))/2))) ϕ^′ (a) =∫_0 ^c ((t(1−t^2 ))/((1+a^2 t^2 )(t^4 +6t^2 +1)))dt =_(at =z) ∫_0 ^(cz) (((z/a)(1−(z^2 /a^2 )))/((1+z^2 )((z^4 /a^4 ) +6(z^2 /a^2 ) +1)))(dz/a) =∫_0 ^(cz) ((z(a^2 −z^2 ))/((1+z^2 )(z^4 +6a^2 z^2 +a^4 ))) dz let decompose F(z) =((z(a^2 −z^2 ))/((z^2 +1)(z^4 +6a^2 z^2 +a^4 ))) z^4 +6a^2 z^2 +a^4 =0→Δ^′ =(3a^2 )^2 −a^4 =8a^4 ⇒ z_1 ^2 =((−3a^2 +2(√2)a^2 )/1) =(2(√2)−3a)a^2 ⇒z_1 =+^− i(√(3a−2(√2))) z_1 ^2 =((−3a^2 −2(√2)a^2 )/1) ⇒z_2 =+^− i(√(3a+2(√2))) .....be continued...
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{arcsin}\left(\mathrm{tan}\theta\right)\mathrm{d}\theta\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\mathrm{arcsin}\left(\mathrm{tan}\theta\right)=\mathrm{x}\:\Rightarrow\mathrm{tan}\theta\:=\mathrm{sinx}\:\Rightarrow\theta\:=\mathrm{artan}\left(\mathrm{sinx}\right)\:\Rightarrow \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\mathrm{tan1}\right)} \:\mathrm{x}\:×\frac{\mathrm{cosx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=_{\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}} \:\int_{\mathrm{0}} ^{\mathrm{tan}\left(\frac{\mathrm{arcsin}\left(\mathrm{tan1}\right)}{\mathrm{2}}\right)} \frac{\mathrm{2arctant}×\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{4t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}×\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{c}} \:\:\:\:\:\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctant}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{4t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{c}} \:\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctant}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{c}} \:\:\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{1}+\mathrm{4t}^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{c}} \:\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{6t}^{\mathrm{2}} +\mathrm{1}}\mathrm{dt} \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{c}} \:\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctan}\left(\mathrm{at}\right)}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{6t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:\:\:\left(\mathrm{c}=\mathrm{tan}\left(\frac{\mathrm{arcsin}\left(\mathrm{tan1}\right)}{\mathrm{2}}\right)\right) \\ $$$$\varphi^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{c}} \:\:\frac{\mathrm{t}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} +\mathrm{6t}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dt} \\ $$$$=_{\mathrm{at}\:=\mathrm{z}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{cz}} \:\:\:\frac{\frac{\mathrm{z}}{\mathrm{a}}\left(\mathrm{1}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\frac{\mathrm{z}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{4}} }\:+\mathrm{6}\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\mathrm{1}\right)}\frac{\mathrm{dz}}{\mathrm{a}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{cz}} \:\:\:\frac{\mathrm{z}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{z}^{\mathrm{4}} \:+\mathrm{6a}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{4}} \right)}\:\mathrm{dz}\:\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{z}\right)\:=\frac{\mathrm{z}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{4}} \:+\mathrm{6a}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{4}} \right)} \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\mathrm{6a}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{4}} \:=\mathrm{0}\rightarrow\Delta^{'} \:=\left(\mathrm{3a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{a}^{\mathrm{4}} \:=\mathrm{8a}^{\mathrm{4}} \:\Rightarrow \\ $$$$\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} =\frac{−\mathrm{3a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\mathrm{a}^{\mathrm{2}} }{\mathrm{1}}\:=\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3a}\right)\mathrm{a}^{\mathrm{2}} \:\Rightarrow\mathrm{z}_{\mathrm{1}} =\overset{−} {+}\mathrm{i}\sqrt{\mathrm{3a}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} \:=\frac{−\mathrm{3a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\mathrm{a}^{\mathrm{2}} }{\mathrm{1}}\:\:\Rightarrow\mathrm{z}_{\mathrm{2}} =\overset{−} {+}\mathrm{i}\sqrt{\mathrm{3a}+\mathrm{2}\sqrt{\mathrm{2}}}\:…..\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$

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