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Question Number 74224 by mathmax by abdo last updated on 20/Nov/19
find ∫ (x^2 +1)^(1/4) cos((1/2)arctan((1/x)))dx and ∫ (x^2 +1)^(1/4) sin((1/2)arctan((1/x)))dx
$${find}\:\int\:\:\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}\:\:{and} \\ $$$$\int\:\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx} \\ $$
Commented by mathmax by abdo last updated on 20/Nov/19
we have x+i =(√(x^2 +1))e^(iarctan((1/x))) ⇒ (√(x+i))=(x^2 +1)^(1/4) e^((i/2)arctan((1/x))) ⇒ ∫(x^2 +1)^(1/4) cos((1/2) arctan((1/x)))dx=Re ( ∫ (√(x+i))dx) and ∫ (x^2 +1)^(1/4) sin((1/2)arctan((1/x)))dx = Im(∫(√(x+i))dx) we have ∫ (√(x+i))dx =_((√(x+i))=t) ∫ t (2t)dt =∫2t^2 dt =(2/3)t^3 +c =(2/3)((√(x+i)))^3 +c =(2/3)(x^2 +1)^(3/4) e^(((3i)/2)arctan((1/x))) +c =(2/3)(x^2 +1)^(3/4) {cos((3/2)arctan((1/x))) +isin((3/2)arctan((1/x))} +c ⇒ ∫ (x^2 +1)^(1/4) cos((1/2)arctan((1/x)))=(2/3)(x^2 +1)^(3/4) cos((3/2)arctan((1/x))) ∫ (x^2 +1)^(1/4) sin((1/2)arctan((1/x)))=(2/3)(x^2 +1)^(3/4) sin((3/2)arctan((1/x)))
$${we}\:{have}\:\:{x}+{i}\:=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$$\sqrt{{x}+{i}}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\:\Rightarrow \\ $$$$\int\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}={Re}\:\left(\:\int\:\:\sqrt{{x}+{i}}{dx}\right)\:{and} \\ $$$$\int\:\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}\:=\:{Im}\left(\int\sqrt{{x}+{i}}{dx}\right)\:{we}\:{have} \\ $$$$\int\:\sqrt{{x}+{i}}{dx}\:=_{\sqrt{{x}+{i}}={t}} \:\:\:\int\:\:{t}\:\:\left(\mathrm{2}{t}\right){dt}\:=\int\mathrm{2}{t}^{\mathrm{2}} {dt}\:=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\sqrt{{x}+{i}}\right)^{\mathrm{3}} \:+{c}\:=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:{e}^{\frac{\mathrm{3}{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \left\{{cos}\left(\frac{\mathrm{3}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:+{isin}\left(\frac{\mathrm{3}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right\}\:+{c}\:\Rightarrow\right. \\ $$$$\int\:\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} {cos}\left(\frac{\mathrm{3}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$$$\int\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:{sin}\left(\frac{\mathrm{3}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$
Commented by mind is power last updated on 20/Nov/19
verry nice sir
$${verry}\:{nice}\:{sir} \\ $$
Answered by mind is power last updated on 20/Nov/19
let u=arctan((1/x))⇒x=(1/(tg(u))) dx=((−(1+tg^2 (u))du)/(tg^2 (u))) ∫(((1+tg^2 (u))/(tg^2 (u))))^(1/4) .cos((u/2)).−(((1+tg^2 (u))/(tg^2 (u)))) du =−∫(((1+tg^2 (u))/(tg^2 (u))))^(5/4) cos((u/2))du ((1+tg^2 (u))/(tg^2 (u)))=(1/(sin^2 (u))) =−∫((1/(sin^2 (u))))^(5/4) .cos((u/2))du =−∫((1/((2sin((u/2))cos((u/2)))^2 )))^(5/4) .cos((u/2))du w=sin((u/2))⇒dw=(1/2)cos((u/2)) ∫((2dw)/({(2w^2 )^(5/4) .(1−w^2 )^(5/4) )) =2^(−(1/4)) ∫(dw/(w.^3 (1−w^2 ) (((1−w^2 )/w^2 ))^(1/4) )) =2^(−(1/4)) ∫((wdw)/(w^4 (1−w^2 )(((1−w^2 )/w^2 ))^(1/4) )) Z=w^2 ⇒2^(−(5/4)) ∫(dz/(z^2 (1−z)(((1−z)/z))^(1/4) )) ((1−z)/z)=u^4 ⇒z=(1/(1+u^4 ))⇒dz=((−4u^3 )/((1+u^4 )^2 )) =−2^(3/2) ∫(u^3 /((1+u^4 )^2 )).(du/(((1/(1+u^4 )))^2 .((u^4 /(1+u^4 ))).u)) =−2^(3/2) ∫((1+u^4 )/u^2 )du=−2^(3/2) (−(1/u)+(u^3 /3))+c
$${let}\:{u}={arctan}\left(\frac{\mathrm{1}}{{x}}\right)\Rightarrow{x}=\frac{\mathrm{1}}{{tg}\left({u}\right)} \\ $$$${dx}=\frac{−\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({u}\right)\right){du}}{{tg}^{\mathrm{2}} \left({u}\right)} \\ $$$$\int\left(\frac{\mathrm{1}+{tg}^{\mathrm{2}} \left({u}\right)}{{tg}^{\mathrm{2}} \left({u}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .{cos}\left(\frac{{u}}{\mathrm{2}}\right).−\left(\frac{\mathrm{1}+{tg}^{\mathrm{2}} \left({u}\right)}{{tg}^{\mathrm{2}} \left({u}\right)}\right)\:{du} \\ $$$$=−\int\left(\frac{\mathrm{1}+{tg}^{\mathrm{2}} \left({u}\right)}{{tg}^{\mathrm{2}} \left({u}\right)}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} {cos}\left(\frac{{u}}{\mathrm{2}}\right){du} \\ $$$$\frac{\mathrm{1}+{tg}^{\mathrm{2}} \left({u}\right)}{{tg}^{\mathrm{2}} \left({u}\right)}=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left({u}\right)} \\ $$$$=−\int\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left({u}\right)}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} .{cos}\left(\frac{{u}}{\mathrm{2}}\right){du} \\ $$$$=−\int\left(\frac{\mathrm{1}}{\left(\mathrm{2}{sin}\left(\frac{{u}}{\mathrm{2}}\right){cos}\left(\frac{{u}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }\right)^{\frac{\mathrm{5}}{\mathrm{4}}} .{cos}\left(\frac{{u}}{\mathrm{2}}\right){du} \\ $$$${w}={sin}\left(\frac{{u}}{\mathrm{2}}\right)\Rightarrow{dw}=\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\frac{{u}}{\mathrm{2}}\right) \\ $$$$\int\frac{\mathrm{2}{dw}}{\left\{\left(\mathrm{2}{w}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{4}}} .\left(\mathrm{1}−{w}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{4}}} \right.} \\ $$$$ \\ $$$$=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \int\frac{{dw}}{{w}.^{\mathrm{3}} \left(\mathrm{1}−{w}^{\mathrm{2}} \right)\:\sqrt[{\mathrm{4}}]{\frac{\mathrm{1}−{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} }}} \\ $$$$=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \int\frac{{wdw}}{{w}^{\mathrm{4}} \left(\mathrm{1}−{w}^{\mathrm{2}} \right)\sqrt[{\mathrm{4}}]{\frac{\mathrm{1}−{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} }}} \\ $$$${Z}={w}^{\mathrm{2}} \Rightarrow\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} \int\frac{{dz}}{{z}^{\mathrm{2}} \left(\mathrm{1}−{z}\right)\sqrt[{\mathrm{4}}]{\frac{\mathrm{1}−{z}}{{z}}}} \\ $$$$\frac{\mathrm{1}−{z}}{{z}}={u}^{\mathrm{4}} \Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }\Rightarrow{dz}=\frac{−\mathrm{4}{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$$$=−\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} }.\frac{{du}}{\left(\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }\right)^{\mathrm{2}} .\left(\frac{{u}^{\mathrm{4}} }{\mathrm{1}+{u}^{\mathrm{4}} }\right).{u}} \\ $$$$=−\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \int\frac{\mathrm{1}+{u}^{\mathrm{4}} }{{u}^{\mathrm{2}} }{du}=−\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(−\frac{\mathrm{1}}{{u}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\right)+{c}\:\:\: \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 20/Nov/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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