Question Number 44781 by arvinddayama01@gmail.com last updated on 04/Oct/18
$$\boldsymbol{{prove}}\:\boldsymbol{{that}}:−\:\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dt}}\:=\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\right)+\boldsymbol{\mathrm{C}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Oct/18
![∫((tdt)/(t^2 (√(1−t^2 )) )) k^2 =1−t^2 2kdk=−2tdt ∫((−kdk)/((1−k^2 )k)) ∫(dk/(k^2 −1)) (1/2)∫(((k+1)−(k−1))/((k+1)(k−1)))dk (1/2)∫(dk/(k−1))−(1/2)∫(dk/(k+1)) (1/2){ln(k−1)−ln(k+1)}+c (1/2){ln(((k−1)/(k+1)))}+c (1/2){ln((((√(1−t^2 )) −1)/( (√(1−t^2 )) +1)))}+c←correct answer recheck of answer given RHS (d/dt){ln(1−(√(1−t^2 )) )} (1/(1−(√(1−t^2 )) ))×((0−(1/(2(√(1−t^2 )) ))×−2t)/1) (1/(1−(√(1−t^2 )) ))×(t/( (√(1−t^2 )))) now recheck result obtained (d/dt)[(1/2){ln((√(1−t^2 )) −1)−ln((√(1−t^2 )) +1)}] (1/2)[{(1/( (√(1−t^2 )) −1))×(1/(2(√(1−t^2 )) ))×−2t}−{(1/( (√(1−t^2 )) +1))×((−2t)/(2(√(1−t^2 )) ))}] (1/2)[((−t)/( (√(1−t^2 )) .((√(1−t^2 )) −1)))+(t/( (√(1−t^2 )) .((√(1−t^2 )) +1)))] (t/(2(√(1−t^2 )) )){((−1)/( (√(1−t_ ^2 )) −1)) +(1/( (√(1−t^2 )) +1)) } (t/(2(√(1−t^2 )))){(((√(1−t)) −1−(√(1−t^2 )) −1)/((1−t^2 −1))} (1/2)×(t/( (√(1−t^2 )))){((−2)/(−t^2 ))} (1/(t(√(1−t^2 )))) so attached answer in right hand side ln(1−(√(1−t^2 )) ) is not correct](https://www.tinkutara.com/question/Q44785.png)
$$\int\frac{{tdt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:} \\ $$$${k}^{\mathrm{2}} =\mathrm{1}−{t}^{\mathrm{2}} \:\:\:\mathrm{2}{kdk}=−\mathrm{2}{tdt} \\ $$$$\int\frac{−{kdk}}{\left(\mathrm{1}−{k}^{\mathrm{2}} \right){k}} \\ $$$$\int\frac{{dk}}{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({k}+\mathrm{1}\right)−\left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)}{dk} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({k}−\mathrm{1}\right)−{ln}\left({k}+\mathrm{1}\right)\right\}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)\right\}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}\right)\right\}+{c}\leftarrow{correct}\:{answer} \\ $$$${recheck}\:{of}\:{answer}\:{given}\:{RHS} \\ $$$$\frac{{d}}{{dt}}\left\{{ln}\left(\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\right)\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×\frac{\mathrm{0}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×−\mathrm{2}{t}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$${now}\:{recheck}\:{result}\:{obtained} \\ $$$$\frac{{d}}{{dt}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}\right)−{ln}\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}\right)\right\}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:−\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:}×−\mathrm{2}{t}\right\}−\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}×\frac{−\mathrm{2}{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}\right\}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:.\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}\right)}+\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:.\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:+\mathrm{1}\right)}\right] \\ $$$$\frac{{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}\left\{\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}_{} ^{\mathrm{2}} }\:\:−\mathrm{1}}\:\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}\:\right\} \\ $$$$\frac{{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left\{\frac{\sqrt{\mathrm{1}−{t}}\:−\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:−\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{1}\right.}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left\{\frac{−\mathrm{2}}{−{t}^{\mathrm{2}} }\right\} \\ $$$$\frac{\mathrm{1}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{attached}}\:\boldsymbol{{answer}}\:\boldsymbol{{in}}\:\boldsymbol{{right}}\:\boldsymbol{{hand}}\:\boldsymbol{{side}} \\ $$$$\boldsymbol{{ln}}\left(\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }\:\:\right)\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{correct}} \\ $$$$ \\ $$
Commented by arvinddayama01@gmail.com last updated on 06/Oct/18
$$\mathrm{thanks} \\ $$
Answered by ajfour last updated on 04/Oct/18
$${let}\:\:{t}=\mathrm{sin}\:\theta \\ $$$$\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}\:=\:\int\mathrm{cosec}\:\theta{d}\theta \\ $$$$=\mathrm{ln}\:\mid\mathrm{cosec}\:\theta−\mathrm{cot}\:\theta\mid+{c} \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{1}}{{t}}−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\mid+{c} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\mid−\mathrm{ln}\:\mid{t}\mid+{c}\:. \\ $$
Commented by arvinddayama01@gmail.com last updated on 06/Oct/18
$$\mathrm{thanks} \\ $$