Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 62209 by maxmathsup by imad last updated on 17/Jun/19

find g(a) =∫(x+a)(√(x^2 −a^2 ))dx

$${find}\:{g}\left({a}\right)\:=\int\left({x}+{a}\right)\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }{dx}\: \\ $$

Commented by maxmathsup by imad last updated on 18/Jun/19

changement x=ach(t) give g(a) =∫ (acht +a)(√(a^2 (ch^2 t−1)))ash(t)dt =a^2 ∣a∣ ∫ (1+ch(t)sh^2 (t) dt =a^2 ∣a∣ ∫2ch^2 t sh^2 t dt =2a^2 ∣a∣ ∫ ((1/2)sh(2t))^2 dt =((a^2 ∣a∣)/2) ∫ ((ch(4t)−1)/2)dt =((a^2 ∣a∣)/4) ∫ (ch(4t)−1)dt =((a^2 ∣a∣)/(16)) sh(4t)−((a^2 ∣a∣)/4) t +c but we have ch(t)=(x/a) ⇒t =argch((x/a))=ln((x/a) +(√((x^2 /a^2 )−1))) ⇒ 4t =ln{ ((x/a) +(√((x^2 /a^2 )−1)))^4 } and sh(4t) =((e^(4t) −e^(−4t) )/2) =(1/2){ ((x/a)+(√((x^2 /a^2 )−1)))^4 −((x/a)+(√((x^2 /a^2 )−1)))^(−4) } ⇒ A =((a^2 ∣a∣)/(32)){ ((x/a) +(√((x^2 /a^2 )−1)))^4 −((x/a)+(√(x^2 /a^2 ))−1)^(−4) }−((a^2 ∣a∣)/4)ln((x/a)+(√((x^2 /a^2 )−1))) +C

$${changement}\:{x}={ach}\left({t}\right)\:{give}\:{g}\left({a}\right)\:=\int\:\:\left({acht}\:+{a}\right)\sqrt{{a}^{\mathrm{2}} \left({ch}^{\mathrm{2}} {t}−\mathrm{1}\right)}{ash}\left({t}\right){dt} \\ $$$$={a}^{\mathrm{2}} \mid{a}\mid\:\int\:\:\left(\mathrm{1}+{ch}\left({t}\right){sh}^{\mathrm{2}} \left({t}\right)\:{dt}\right. \\ $$$$={a}^{\mathrm{2}} \mid{a}\mid\:\int\mathrm{2}{ch}^{\mathrm{2}} {t}\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\mathrm{2}{a}^{\mathrm{2}} \mid{a}\mid\:\int\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} {dt} \\ $$$$=\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{2}}\:\int\:\:\:\frac{{ch}\left(\mathrm{4}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\:=\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{4}}\:\int\:\:\left({ch}\left(\mathrm{4}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{16}}\:{sh}\left(\mathrm{4}{t}\right)−\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{4}}\:{t}\:+{c}\:\:{but}\:{we}\:{have}\: \\ $$$${ch}\left({t}\right)=\frac{{x}}{{a}}\:\Rightarrow{t}\:={argch}\left(\frac{{x}}{{a}}\right)={ln}\left(\frac{{x}}{{a}}\:+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)\:\Rightarrow \\ $$$$\mathrm{4}{t}\:={ln}\left\{\:\left(\frac{{x}}{{a}}\:+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)^{\mathrm{4}} \right\}\:\:{and}\: \\ $$$${sh}\left(\mathrm{4}{t}\right)\:=\frac{{e}^{\mathrm{4}{t}} \:−{e}^{−\mathrm{4}{t}} }{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left(\frac{{x}}{{a}}+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)^{−\mathrm{4}} \right\}\:\Rightarrow \\ $$$${A}\:=\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{32}}\left\{\:\left(\frac{{x}}{{a}}\:+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)^{−\mathrm{4}} \right\}−\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{4}}{ln}\left(\frac{{x}}{{a}}+\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}\right)\:+{C}\: \\ $$

Answered by tanmay last updated on 17/Jun/19

∫x(√(x^2 −a^2 )) dx+a∫(√(x^2 −a^2 )) dx (1/2)∫(x^2 −a^2 )^(1/2) d(x^2 −a^2 )+a∫(√(x^2 −a^2 )) dx (1/2)×(((x^2 −a^2 )^(3/2) )/(3/2))+a[((x(√(x^2 −a^2 )))/2)−(a^2 /2)ln(x+(√(x^2 −a^2 )) )]+c

$$\int{x}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:{dx}+{a}\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {d}\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{a}\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}+{a}\left[\frac{{x}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\right)\right]+{c} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com