Question Number 179513 by cortano1 last updated on 30/Oct/22
$$\:\mathrm{F}\left(\mathrm{x}\right)\:=\:\int\:\frac{\mathrm{1}}{\mathrm{x}}\:\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\:\mathrm{dx}\: \\ $$$$\:\:\mathrm{F}\left(\mathrm{1}\right)=\mathrm{0}\: \\ $$$$\:\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$
Answered by mr W last updated on 30/Oct/22
![let t=(√((1−x)/(1+x))) x=((1−t^2 )/(1+t^2 )) dx=[((−2t)/(1+t^2 ))−((2t(1−t^2 ))/((1+t^2 )^2 ))]dt=((−4t)/((1+t^2 )^2 ))dt F(x)=∫(((1+t^2 )/(1−t^2 )))t((−4t)/((1+t^2 )^2 ))dt =∫((−4t^2 )/((1+t^2 )(1−t^2 )))dt =2∫((1/(1+t^2 ))−(1/(1−t^2 )))dt =2 tan^(−1) t−ln ((1+t)/(1−t))+C =2 tan^(−1) (√((1−x)/(1+x)))−ln (((√(1+x))+(√(1−x)))/( (√(1+x))−(√(1−x))))+C F(1)=2 tan^(−1) (√(0/2))−ln ((√2)/( (√2)))+C=0 ⇒C=0 F(x)=2 tan^(−1) (√((1−x)/(1+x)))−ln (((√(1+x))+(√(1−x)))/( (√(1+x))−(√(1−x)))) F((1/2))=2 tan^(−1) (1/( (√3)))−ln (((√3)+1)/( (√3)−1)) F((1/2))=(π/3)−ln (2+(√3)) =(π/3)+ln (2−(√3)) ✓](https://www.tinkutara.com/question/Q179521.png)
$${let}\:{t}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}} \\ $$$${x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${dx}=\left[\frac{−\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]{dt}=\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$${F}\left({x}\right)=\int\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right){t}\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\:\:\:=\int\frac{−\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\:\:\:=\mathrm{2}\int\left(\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\right){dt} \\ $$$$\:\:\:=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} {t}−\mathrm{ln}\:\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}+{C} \\ $$$$\:\:\:=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}−\mathrm{ln}\:\frac{\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}+{C} \\ $$$$\:{F}\left(\mathrm{1}\right)=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{0}}{\mathrm{2}}}−\mathrm{ln}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}+{C}=\mathrm{0} \\ $$$$\:\Rightarrow{C}=\mathrm{0} \\ $$$${F}\left({x}\right)=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}−\mathrm{ln}\:\frac{\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}} \\ $$$${F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{ln}\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}} \\ $$$${F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{3}}−\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:=\frac{\pi}{\mathrm{3}}+\mathrm{ln}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\checkmark \\ $$
Commented by cortano1 last updated on 30/Oct/22
$$\:\mathrm{i}\:\mathrm{got}\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{3}}+\mathrm{ln}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\: \\ $$
Commented by Tawa11 last updated on 30/Oct/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 30/Oct/22
$${thanks}\:{for}\:{understanding}! \\ $$
Commented by ARUNG_Brandon_MBU last updated on 30/Oct/22
��The reason for the "Great Sir" on all posts according to me is for them to be saved in "My Post" section so they can easily be found ��
Miss Tawa, instead of using "Great Sir" to save posts you can instead use the bookmark option���� unless it's not what I think.
Commented by mr W last updated on 30/Oct/22
$${i}'{ll}\:{ask}\:{tinku}\:{tara}\:{to}\:{add}\:{the}\:{feature} \\ $$$${that}\:{one}\:{receives}\:{notifications}\:{also} \\ $$$${when}\:{his}\:{bookmarked}\:{posts}\:{get}\: \\ $$$${updated}. \\ $$
Commented by Tawa11 last updated on 30/Oct/22
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{Mr}\:\mathrm{brandon}\:\mathrm{got}\:\mathrm{it}. \\ $$$$\mathrm{I}\:\mathrm{like}\:\mathrm{to}\:\mathrm{save}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{future}\:\mathrm{purpose}. \\ $$
Commented by Tawa11 last updated on 30/Oct/22
$$\mathrm{Read}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{get}\:\mathrm{your}\:\mathrm{point}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 30/Oct/22
$${i}\:{understood}\:{that}\:{you}\:{just}\:{want}\:{to}\: \\ $$$${track}\:{the}\:{posts}\:{in}\:{this}\:{way},\:{instead}\:{of} \\ $$$${really}\:{commenting}\:{the}\:{posts}. \\ $$$${bookmarking}\:{the}\:{posts}\:{is}\:{a}\:{better}\:{way}. \\ $$$${but}\:{you}\:{won}'{t}\:{be}\:{notificated}\:{when}\:{the} \\ $$$${posts}\:{are}\:{updated}.\:{therefore}\:{i}\:{want}\:{to} \\ $$$${asl}\:{tinku}\:{tara}\:{adding}\:{a}\:{new}\:{feature}, \\ $$$${that}\:{one}\:{gets}\:{notifications}\:{when}\:{the} \\ $$$${bookmarked}\:{posted}\:{get}\:{updated}. \\ $$
Commented by Rasheed.Sindhi last updated on 30/Oct/22
$$\mathrm{And}\:\mathrm{we}\:\mathrm{be}\:\mathrm{happy}\:\mathrm{that}\:\mathrm{miss}\:\mathrm{tawa}\:\mathrm{has} \\ $$$$\mathrm{admired}\:\mathrm{us}!!!\:{hahaha}…\: \\ $$
Commented by Ar Brandon last updated on 30/Oct/22
������
Commented by Tawa11 last updated on 31/Oct/22
$$\mathrm{Sir},\:\mathrm{tagging}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{save}\:\mathrm{it}. \\ $$$$\mathrm{Tagging}\:\mathrm{not}\:\mathrm{to}\:\mathrm{loose}\:\mathrm{your}\:\mathrm{solutions}. \\ $$$$\mathrm{Trying}\:\mathrm{my}\:\mathrm{own}\:\mathrm{way}\:\mathrm{to}\:\mathrm{save}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{evidence}\:\mathrm{that}\:\mathrm{I}\:\mathrm{admire}\:\mathrm{your}\:\mathrm{solutions}. \\ $$