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Question Number 59576 by maxmathsup by imad last updated on 12/May/19
let f(x) =∫ (dt/((x+t)(√(t^2 −x^2 )))) 1) determine a explicit form of f(x) 2) determine ∫ (dt/((x+2)(√(t^2 −4)))) and ∫ (dt/((x+1)(√(t^2 −1))))
$${let}\:{f}\left({x}\right)\:=\int\:\:\:\:\:\:\:\frac{{dt}}{\left({x}+{t}\right)\sqrt{{t}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:\int\:\:\:\:\:\frac{{dt}}{\left({x}+\mathrm{2}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:\:{and}\:\:\int\:\:\:\:\:\:\frac{{dt}}{\left({x}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$
Commented by tanmay last updated on 12/May/19
∫(dt/((t+2)(√(t^2 −4)))) and ∫(dt/((t+1)(√(t^2 −1)))) ∫(dt/((t+a)(√(t^2 −a^2 )))) k=(1/(t+a)) t+a=(1/k)→dt=((−dk)/k^2 ) ∫(k/( (√(((1/k))((1/k)−2a)))))×((−dk)/k^2 ) ∫((−dk)/(k(√((1−2ak)/k^2 )))) p^2 =1−2ak →2pdp=−2adk ∫((pdp)/(a×p)) (1/a)×p+c =((√(1−2ak))/a)+c =((√(1−((2a)/(t+a)) ))/a)+c now put a=2 ((√(1−(4/(t+2))))/2)+c....(ans for Qno 1) now put a=1 ((√(1−(2/(t+1))))/1)+c....(ans for Qno 2)
$$\int\frac{{dt}}{\left({t}+\mathrm{2}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:{and}\:\int\frac{{dt}}{\left({t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\int\frac{{dt}}{\left({t}+{a}\right)\sqrt{{t}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${k}=\frac{\mathrm{1}}{{t}+{a}} \\ $$$${t}+{a}=\frac{\mathrm{1}}{{k}}\rightarrow{dt}=\frac{−{dk}}{{k}^{\mathrm{2}} } \\ $$$$\int\frac{{k}}{\:\sqrt{\left(\frac{\mathrm{1}}{{k}}\right)\left(\frac{\mathrm{1}}{{k}}−\mathrm{2}{a}\right)}}×\frac{−{dk}}{{k}^{\mathrm{2}} } \\ $$$$\int\frac{−{dk}}{{k}\sqrt{\frac{\mathrm{1}−\mathrm{2}{ak}}{{k}^{\mathrm{2}} }}}\: \\ $$$${p}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{ak}\:\:\:\rightarrow\mathrm{2}{pdp}=−\mathrm{2}{adk} \\ $$$$\int\frac{{pdp}}{{a}×{p}} \\ $$$$\frac{\mathrm{1}}{{a}}×{p}+{c} \\ $$$$=\frac{\sqrt{\mathrm{1}−\mathrm{2}{ak}}}{{a}}+{c} \\ $$$$=\frac{\sqrt{\mathrm{1}−\frac{\mathrm{2}{a}}{{t}+{a}}\:}}{{a}}+{c} \\ $$$${now}\:{put}\:{a}=\mathrm{2}\:\: \\ $$$$\frac{\sqrt{\mathrm{1}−\frac{\mathrm{4}}{{t}+\mathrm{2}}}}{\mathrm{2}}+{c}….\left({ans}\:{for}\:{Qno}\:\mathrm{1}\right) \\ $$$${now}\:{put}\:{a}=\mathrm{1} \\ $$$$\frac{\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{t}+\mathrm{1}}}}{\mathrm{1}}+{c}….\left({ans}\:{for}\:{Qno}\:\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 12/May/19
2)the Q.is determine ∫ (dt/((t+2)(√(t^2 −4)))) and ∫ (dt/((t+1)(√(t^2 −1))))
$$\left.\mathrm{2}\right){the}\:{Q}.{is}\:{determine}\:\int\:\:\frac{{dt}}{\left({t}+\mathrm{2}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:\:{and}\:\int\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 14/May/19
changement t =x ch(u) give f(x)=∫ ((xsh(u))/((x+xch(u))∣x∣sh(u))) du =ξ(x) ∫ (du/(x(1+ch(u)))) =((ξ(x))/x) ∫ (du/(1+((e^u +e^(−u) )/2))) =((2ξ(x))/x) ∫ (du/(2 +e^u +e^(−u) )) =_(e^u =α) ((2ξ(x))/x) ∫ (1/(2 +α +α^(−1) )) (dα/α) =((2ξ(x))/x) ∫ (dα/(2α +α^2 +1)) =((2ξ(x))/x) ∫ (dα/((α+1)^2 )) =−((2ξ(x))/x) (1/(α+1)) +c =−(2/x)ξ(x) (1/(e^u +1)) but u =argch((t/x)) =ln((t/x)+(√((t^2 /x^2 )−1))) ⇒e^u =(t/x)+(√((t^2 /x^2 )−1)) ⇒ f(x) =−(2/x)ξ(x) (1/((t/x)+((√(t^2 −x^2 ))/(∣x∣)))) +c =−((2ξ(x))/(t +ξ(x)(√(t^2 −x^2 )))) +c with ξ(x) =1 if x>0 and ξ(x)=−1 if x<0 so f(x) =((−2)/(t+(√(t^2 −x^2 )))) +c if x>0 f(x)=(2/(t−(√(t^2 −x^2 ))))+c if x<0
$${changement}\:{t}\:={x}\:{ch}\left({u}\right)\:{give}\:{f}\left({x}\right)=\int\:\:\frac{{xsh}\left({u}\right)}{\left({x}+{xch}\left({u}\right)\right)\mid{x}\mid{sh}\left({u}\right)}\:{du} \\ $$$$=\xi\left({x}\right)\:\int\:\:\:\frac{{du}}{{x}\left(\mathrm{1}+{ch}\left({u}\right)\right)}\:=\frac{\xi\left({x}\right)}{{x}}\:\int\:\:\:\frac{{du}}{\mathrm{1}+\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}\:=\frac{\mathrm{2}\xi\left({x}\right)}{{x}}\:\int\:\:\:\frac{{du}}{\mathrm{2}\:+{e}^{{u}} \:+{e}^{−{u}} } \\ $$$$=_{{e}^{{u}} =\alpha} \frac{\mathrm{2}\xi\left({x}\right)}{{x}}\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{2}\:+\alpha\:+\alpha^{−\mathrm{1}} }\:\frac{{d}\alpha}{\alpha}\:=\frac{\mathrm{2}\xi\left({x}\right)}{{x}}\:\int\:\:\:\:\frac{{d}\alpha}{\mathrm{2}\alpha\:+\alpha^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\xi\left({x}\right)}{{x}}\:\int\:\:\frac{{d}\alpha}{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{2}\xi\left({x}\right)}{{x}}\:\frac{\mathrm{1}}{\alpha+\mathrm{1}}\:+{c}\:=−\frac{\mathrm{2}}{{x}}\xi\left({x}\right)\:\frac{\mathrm{1}}{{e}^{{u}} \:+\mathrm{1}}\:\:{but} \\ $$$${u}\:={argch}\left(\frac{{t}}{{x}}\right)\:={ln}\left(\frac{{t}}{{x}}+\sqrt{\frac{{t}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\mathrm{1}}\right)\:\Rightarrow{e}^{{u}} \:=\frac{{t}}{{x}}+\sqrt{\frac{{t}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=−\frac{\mathrm{2}}{{x}}\xi\left({x}\right)\:\frac{\mathrm{1}}{\frac{{t}}{{x}}+\frac{\sqrt{{t}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mid{x}\mid}}\:+{c}\:=−\frac{\mathrm{2}\xi\left({x}\right)}{{t}\:+\xi\left({x}\right)\sqrt{{t}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:+{c}\:\:{with} \\ $$$$\xi\left({x}\right)\:=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:\:{and}\:\xi\left({x}\right)=−\mathrm{1}\:{if}\:{x}<\mathrm{0}\:{so}\: \\ $$$${f}\left({x}\right)\:=\frac{−\mathrm{2}}{{t}+\sqrt{{t}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:+{c}\:\:{if}\:{x}>\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{{t}−\sqrt{{t}^{\mathrm{2}} −{x}^{\mathrm{2}} }}+{c}\:{if}\:{x}<\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 14/May/19
2) ∫ (dt/((t+2)(√(t^2 −4)))) =f(2) =((−2)/(t +(√(t^2 −4)))) +c ∫ (dt/((t+1)(√(t^2 −1)))) =f(1) =((−2)/(t +(√(t^2 −1)))) +c
$$\left.\mathrm{2}\right)\:\int\:\:\:\frac{{dt}}{\left({t}+\mathrm{2}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:={f}\left(\mathrm{2}\right)\:=\frac{−\mathrm{2}}{{t}\:+\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}\:+{c} \\ $$$$\int\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:={f}\left(\mathrm{1}\right)\:=\frac{−\mathrm{2}}{{t}\:+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:+{c}\: \\ $$

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