Question Number 85909 by jagoll last updated on 26/Mar/20
$$\int\:\mathrm{sin}^{−\mathrm{1}} \:\left(\sqrt{\frac{\mathrm{x}}{\mathrm{a}+\mathrm{x}}}\right)\:\mathrm{dx}\:,\:\mathrm{a}\:>\:\mathrm{0} \\ $$
Commented by john santu last updated on 26/Mar/20
![let (√(x/(a+x))) = n ⇒ x = ((an^2 )/(1−n^2 )) I = ∫ sin^(−1) (n) × ((2an)/((1−n^2 )^2 )) dn [ h = sin^(−1) (n) ⇒ n = sin h ] I = 2a∫ h tan h sec^2 h dh I = a ( h tan^2 h −∫ (sec^2 h −1)dh) I = a (h tan^2 h −tan h + h) +c ∴ I = x sin^(−1) ((√(x/(a+x))))−(√(ax)) + a sin^(−1) ((√(x/(a+x)))) + c](https://www.tinkutara.com/question/Q85947.png)
$${let}\:\sqrt{\frac{{x}}{{a}+{x}}}\:=\:{n}\:\Rightarrow\:{x}\:=\:\frac{{an}^{\mathrm{2}} }{\mathrm{1}−{n}^{\mathrm{2}} } \\ $$$${I}\:=\:\int\:\mathrm{sin}^{−\mathrm{1}} \left({n}\right)\:×\:\frac{\mathrm{2}{an}}{\left(\mathrm{1}−{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dn} \\ $$$$\left[\:{h}\:=\:\mathrm{sin}^{−\mathrm{1}} \left({n}\right)\:\Rightarrow\:{n}\:=\:\mathrm{sin}\:{h}\:\right] \\ $$$${I}\:=\:\mathrm{2}{a}\int\:{h}\:\mathrm{tan}\:{h}\:\mathrm{sec}\:^{\mathrm{2}} {h}\:{dh}\: \\ $$$${I}\:=\:{a}\:\left(\:{h}\:\mathrm{tan}\:^{\mathrm{2}} {h}\:−\int\:\left(\mathrm{sec}\:^{\mathrm{2}} {h}\:−\mathrm{1}\right){dh}\right) \\ $$$${I}\:=\:{a}\:\left({h}\:\mathrm{tan}\:^{\mathrm{2}} {h}\:−\mathrm{tan}\:{h}\:+\:{h}\right)\:+{c} \\ $$$$\therefore\:{I}\:=\:{x}\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{{a}+{x}}}\right)−\sqrt{{ax}}\:+\:{a}\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{{a}+{x}}}\right)\:+\:{c} \\ $$
Commented by jagoll last updated on 26/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all} \\ $$
Answered by TANMAY PANACEA. last updated on 26/Mar/20
![x=atan^2 α→(dx/dα)=a×2tanα×sec^2 α (√((atan^2 α)/(a+atan^2 α))) =(√((sin^2 α)/(cos^2 α×sec^2 α))) =sinα I=∫α.2atanα.sec^2 αdα (I/(2a))=α∫tanαsec^2 αdα−∫[(dα/dα)∫tanαsec^2 αdα]dα =α.((tan^2 α)/2)−(1/2)∫(sec^2 α−1)dα =((αtan^2 α)/2)−((tanα)/2)+(α/2) now α=tan^(−1) (√(x/a)) I=(((tan^(−1) (√(x/a)) ×(x/a))/2)−((√(x/a))/2)+((tan^(−1) (√(x/a)))/2))×2a I=xtan^(−1) (√(x/a)) −a(√(x/a)) +atan^(−1) (√(x/a)) =(x+a)tan^(−1) (√(x/a)) −(√(ax)) pls check](https://www.tinkutara.com/question/Q85928.png)
$${x}={atan}^{\mathrm{2}} \alpha\rightarrow\frac{{dx}}{{d}\alpha}={a}×\mathrm{2}{tan}\alpha×{sec}^{\mathrm{2}} \alpha \\ $$$$\sqrt{\frac{{atan}^{\mathrm{2}} \alpha}{{a}+{atan}^{\mathrm{2}} \alpha}}\:=\sqrt{\frac{{sin}^{\mathrm{2}} \alpha}{{cos}^{\mathrm{2}} \alpha×{sec}^{\mathrm{2}} \alpha}}\:={sin}\alpha \\ $$$${I}=\int\alpha.\mathrm{2}{atan}\alpha.{sec}^{\mathrm{2}} \alpha{d}\alpha \\ $$$$\frac{{I}}{\mathrm{2}{a}}=\alpha\int{tan}\alpha{sec}^{\mathrm{2}} \alpha{d}\alpha−\int\left[\frac{{d}\alpha}{{d}\alpha}\int{tan}\alpha{sec}^{\mathrm{2}} \alpha{d}\alpha\right]{d}\alpha \\ $$$$=\alpha.\frac{{tan}^{\mathrm{2}} \alpha}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\left({sec}^{\mathrm{2}} \alpha−\mathrm{1}\right){d}\alpha \\ $$$$=\frac{\alpha{tan}^{\mathrm{2}} \alpha}{\mathrm{2}}−\frac{{tan}\alpha}{\mathrm{2}}+\frac{\alpha}{\mathrm{2}} \\ $$$${now}\:\alpha={tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\: \\ $$$${I}=\left(\frac{{tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\:×\frac{{x}}{{a}}}{\mathrm{2}}−\frac{\sqrt{\frac{{x}}{{a}}}}{\mathrm{2}}+\frac{{tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}}{\mathrm{2}}\right)×\mathrm{2}{a} \\ $$$${I}={xtan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\:−{a}\sqrt{\frac{{x}}{{a}}}\:+{atan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\: \\ $$$$=\left({x}+{a}\right){tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\:−\sqrt{{ax}}\: \\ $$$${pls}\:{check} \\ $$
Answered by Kunal12588 last updated on 26/Mar/20
$${I}={xsin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}−\int{x}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{{x}}{{a}+{x}}}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{{x}}{{a}+{x}}}}×\frac{{a}+{x}−{x}}{\left({a}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$={xsin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}−\int{x}×\frac{\sqrt{{a}+{x}}}{\:\sqrt{{a}}}×\frac{\sqrt{{a}+{x}}}{\mathrm{2}\sqrt{{x}}}×\frac{{a}}{\left({a}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$={xsin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}}\int\sqrt{{x}}×\frac{\mathrm{1}}{{a}+{x}}{dx} \\ $$$${I}_{\mathrm{1}} =\int\frac{\sqrt{{x}}}{{a}+{x}}{dx} \\ $$$$\sqrt{{x}}={t} \\ $$$${dx}=\mathrm{2}\sqrt{{x}}{dt} \\ $$$${I}_{\mathrm{1}} =\mathrm{2}\int\frac{{x}}{{a}+{x}}{dt}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{a}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}\int\frac{{a}+{t}^{\mathrm{2}} }{{a}+{t}^{\mathrm{2}} }{dt}−\mathrm{2}{a}\int\frac{{dt}}{{a}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}{t}−\mathrm{2}{a}×\frac{\mathrm{1}}{\:\sqrt{{a}}}{tan}^{−\mathrm{1}} \frac{{t}}{\:\sqrt{{a}}}+{c} \\ $$$$=\mathrm{2}\sqrt{{x}}−\mathrm{2}\sqrt{{a}}\:{tan}^{−\mathrm{1}} \frac{\sqrt{{x}}}{\:\sqrt{{a}}}+{c} \\ $$$${I}={xsin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}}\left(\mathrm{2}\sqrt{{x}}−\mathrm{2}\sqrt{{a}}\:{tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}\right)+{C} \\ $$$${I}={xsin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}−\sqrt{{ax}}+{a}\:{tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}}+{C} \\ $$
Commented by Kunal12588 last updated on 26/Mar/20
$${sin}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}+{x}}}={tan}^{−\mathrm{1}} \sqrt{\frac{{x}}{{a}}} \\ $$$${so}\:{Tanmay}\:{sir}\:{your}\:{answer}\:{is}\:{also}\:{correct}! \\ $$