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Question Number 55274 by maxmathsup by imad last updated on 20/Feb/19
let ϕ(a) =∫_1 ^(√3) arctan((a/x))dx 1) calculate ϕ(a) interms of a 2) calculate ϕ^′ (a) at form of integral. 3) determine ϕ^((n)) (a) at form of integral. 4) find the value of ∫_1 ^(√3) arctan((2/x))dx .
$${let}\:\varphi\left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{arctan}\left(\frac{{a}}{{x}}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi\left({a}\right)\:{interms}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\varphi^{'} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{3}\right)\:{determine}\:\varphi^{\left({n}\right)} \left({a}\right)\:\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:{arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 24/Feb/19
1) changement (a/x) =t give x =(a/t) ⇒ϕ(a)=∫_a ^(a/( (√3))) arctan(t)(((−a)/t^2 ))dt ⇒ ((ϕ(a))/a)= ∫_(a/( (√3))) ^a ((arctan(t))/t^2 ) dt by parts we get ((ϕ(a))/a) =[−(1/t) arctan(t)]_(a/( (√3))) ^a +∫_(a/( (√3))) ^a (1/(t(1+t^2 )))dt =((√3)/a) arctan((a/( (√3)))) −((arctan(a))/a) +∫_(a/( (√3))) ^a (dt/(t(1+t^2 )))dt but ∫_(a/( (√3))) ^a (dt/(t(1+t^2 )))dt =∫_(a/( (√3))) ^a ((1/t) −(t/(1+t^2 )))dt = [ln((t/( (√(1+t^2 )))))]_(a/( (√3))) ^a =ln((a/( (√(1+a^2 ))))) −ln((a/( (√3)((√(1+(a^2 /3))))))=ln((a/( (√(1+a^2 )))))−ln((a/( (√(a^2 +3))))) =ln(a)−(1/2)ln(1+a^2 )−ln(a)+(1/2)ln(a^2 +3) =(1/2){ln(a^2 +3)−ln(a^2 +1)} ⇒ ((ϕ(a))/a) =((√3)/a) arctan((a/( (√3))))−((arctan(a))/a) +ln((√((a^2 +3)/(a^2 +1)))) ⇒ ϕ(a) =(√3)arctan((a/( (√3))))−arctan(a) +aln((√((a^2 +3)/(a^2 +1)))) .
$$\left.\mathrm{1}\right)\:{changement}\:\frac{{a}}{{x}}\:={t}\:{give}\:{x}\:=\frac{{a}}{{t}}\:\Rightarrow\varphi\left({a}\right)=\int_{{a}} ^{\frac{{a}}{\:\sqrt{\mathrm{3}}}} \:\:\:{arctan}\left({t}\right)\left(\frac{−{a}}{{t}^{\mathrm{2}} }\right){dt}\:\Rightarrow \\ $$$$\frac{\varphi\left({a}\right)}{{a}}=\:\int_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:\:\:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} }\:{dt}\:\:{by}\:{parts}\:{we}\:{get}\:\frac{\varphi\left({a}\right)}{{a}}\:=\left[−\frac{\mathrm{1}}{{t}}\:{arctan}\left({t}\right)\right]_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:+\int_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{{a}}\:{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{3}}}\right)\:−\frac{{arctan}\left({a}\right)}{{a}}\:+\int_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:\:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:{but} \\ $$$$\int_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\int_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} \:\left(\frac{\mathrm{1}}{{t}}\:−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt}\:=\:\left[{ln}\left(\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\right)\right]_{\frac{{a}}{\:\sqrt{\mathrm{3}}}} ^{{a}} ={ln}\left(\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right) \\ $$$$−{ln}\left(\frac{{a}}{\:\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}\right.}\right)={ln}\left(\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right)−{ln}\left(\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}}\right) \\ $$$$={ln}\left({a}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)−{ln}\left({a}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({a}^{\mathrm{2}} \:+\mathrm{3}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({a}^{\mathrm{2}} +\mathrm{3}\right)−{ln}\left({a}^{\mathrm{2}} \:+\mathrm{1}\right)\right\}\:\Rightarrow \\ $$$$\frac{\varphi\left({a}\right)}{{a}}\:=\frac{\sqrt{\mathrm{3}}}{{a}}\:{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{3}}}\right)−\frac{{arctan}\left({a}\right)}{{a}}\:+{ln}\left(\sqrt{\frac{{a}^{\mathrm{2}} +\mathrm{3}}{{a}^{\mathrm{2}} +\mathrm{1}}}\right)\:\:\Rightarrow \\ $$$$\varphi\left({a}\right)\:=\sqrt{\mathrm{3}}{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left({a}\right)\:+{aln}\left(\sqrt{\frac{{a}^{\mathrm{2}} \:+\mathrm{3}}{{a}^{\mathrm{2}} \:+\mathrm{1}}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 24/Feb/19
2) we have ϕ^′ (a) =∫_1 ^(√3) (1/(x(1+(a^2 /x^2 )))) dx = ∫_1 ^(√3) (x/(x^2 +a^2 )) dx .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\varphi^{'} \left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}}{{x}\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:{dx}\:=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:{dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 24/Feb/19
4) we have ϕ(a) =∫_1 ^(√3) arctan((a/x))dx ⇒ ∫_1 ^(√3) arctan((2/x))dx =ϕ(2) =(√3)arctan((2/( (√3))))−arctan(2) +2 ln((√(7/5))) =(√3)arctan((2/( (√3))))−arctan(2) +ln(7)−ln(5) .
$$\left.\mathrm{4}\right)\:{we}\:{have}\:\varphi\left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{arctan}\left(\frac{{a}}{{x}}\right){dx}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:=\varphi\left(\mathrm{2}\right)\:=\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\mathrm{2}\right)\:+\mathrm{2}\:{ln}\left(\sqrt{\frac{\mathrm{7}}{\mathrm{5}}}\right) \\ $$$$=\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\mathrm{2}\right)\:+{ln}\left(\mathrm{7}\right)−{ln}\left(\mathrm{5}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 24/Feb/19
3) we have ϕ^((1)) (a) =∫_1 ^(√3) (x/(x^2 +a^2 ))dx ⇒ϕ^((n)) (a) =∫_1 ^(√3) (d^(n−1) /da^(n−1) ) ((x/(x^2 +a^2 ))) dx let =∫_1 ^(√3) x {(1/(x^2 +a^2 ))}_(/a) ^((n−1)) dx =∫_1 ^(√3) (1/(2i)) {(1/(a−ix)) −(1/(a+ix))}_(/a) ^((n−1)) dx =(1/(2i)) ∫_1 ^(√3) { (((−1)^(n−1) (n−1)!)/((a−ix)^n )) −(((−1)^(n−1) (n−1)!)/((a+ix)^n ))}dx =(((−1)^(n−1) (n−1)!)/(2i)) ∫_1 ^(√3) (((a+ix)^n −(a−ix)^n )/((a^2 +x^2 )^n )) dx =(−1)^(n−1) (n−1)! ∫_1 ^(√3) ((Im((a+ix)^n ))/((a^2 +x^2 )^n )) dx .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\varphi^{\left(\mathrm{1}\right)} \left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{x}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\Rightarrow\varphi^{\left({n}\right)} \left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \frac{{d}^{{n}−\mathrm{1}} }{{da}^{{n}−\mathrm{1}} }\:\:\left(\frac{{x}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\right)\:{dx}\:\:{let} \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{x}\:\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\right\}_{/{a}} ^{\left({n}−\mathrm{1}\right)} \:\:{dx}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left\{\frac{\mathrm{1}}{{a}−{ix}}\:−\frac{\mathrm{1}}{{a}+{ix}}\right\}_{/{a}} ^{\left({n}−\mathrm{1}\right)} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\:\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({a}−{ix}\right)^{{n}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({a}+{ix}\right)^{{n}} }\right\}{dx} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\frac{\left({a}+{ix}\right)^{{n}} −\left({a}−{ix}\right)^{{n}} }{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{{n}} }\:{dx} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\:\frac{{Im}\left(\left({a}+{ix}\right)^{{n}} \right)}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{{n}} }\:{dx}\:. \\ $$

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