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Question Number 44781 by arvinddayama01@gmail.com last updated on 04/Oct/18

prove that:− ∫(1/(t(√(1−t^2 ))))dt = ln(1−(√(1−t^2 )))+C

$$\boldsymbol{{prove}}\:\boldsymbol{{that}}:−\:\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dt}}\:=\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\right)+\boldsymbol{\mathrm{C}} \\ $$

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

$$\int\frac{{tdt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:} \\ $$$${k}^{\mathrm{2}} =\mathrm{1}−{t}^{\mathrm{2}} \:\:\:\mathrm{2}{kdk}=−\mathrm{2}{tdt} \\ $$$$\int\frac{−{kdk}}{\left(\mathrm{1}−{k}^{\mathrm{2}} \right){k}} \\ $$$$\int\frac{{dk}}{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({k}+\mathrm{1}\right)−\left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)}{dk} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({k}−\mathrm{1}\right)−{ln}\left({k}+\mathrm{1}\right)\right\}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)\right\}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}\right)\right\}+{c}\leftarrow{correct}\:{answer} \\ $$$${recheck}\:{of}\:{answer}\:{given}\:{RHS} \\ $$$$\frac{{d}}{{dt}}\left\{{ln}\left(\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\right)\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×\frac{\mathrm{0}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×−\mathrm{2}{t}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}×\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$${now}\:{recheck}\:{result}\:{obtained} \\ $$$$\frac{{d}}{{dt}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}\right)−{ln}\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}\right)\right\}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:−\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:}×−\mathrm{2}{t}\right\}−\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}×\frac{−\mathrm{2}{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}\right\}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:.\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:−\mathrm{1}\right)}+\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:.\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:+\mathrm{1}\right)}\right] \\ $$$$\frac{{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}\left\{\frac{−\mathrm{1}}{\sqrt{\mathrm{1}−{t}_{} ^{\mathrm{2}} }\:\:−\mathrm{1}}\:\:+\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{1}}\:\right\} \\ $$$$\frac{{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left\{\frac{\sqrt{\mathrm{1}−{t}}\:−\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:−\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{1}\right.}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left\{\frac{−\mathrm{2}}{−{t}^{\mathrm{2}} }\right\} \\ $$$$\frac{\mathrm{1}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{attached}}\:\boldsymbol{{answer}}\:\boldsymbol{{in}}\:\boldsymbol{{right}}\:\boldsymbol{{hand}}\:\boldsymbol{{side}} \\ $$$$\boldsymbol{{ln}}\left(\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }\:\:\right)\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{correct}} \\ $$$$ \\ $$

Commented by arvinddayama01@gmail.com last updated on 06/Oct/18

thanks

$$\mathrm{thanks} \\ $$

Answered by ajfour last updated on 04/Oct/18

let  t=sin θ  ∫((cos θdθ)/(sin θcos θ)) = ∫cosec θdθ  =ln ∣cosec θ−cot θ∣+c  =ln ∣(1/t)−((√(1−t^2 ))/t)∣+c  = ln ∣1−(√(1−t^2 )) ∣−ln ∣t∣+c .

$${let}\:\:{t}=\mathrm{sin}\:\theta \\ $$$$\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}\:=\:\int\mathrm{cosec}\:\theta{d}\theta \\ $$$$=\mathrm{ln}\:\mid\mathrm{cosec}\:\theta−\mathrm{cot}\:\theta\mid+{c} \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{1}}{{t}}−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\mid+{c} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\mid−\mathrm{ln}\:\mid{t}\mid+{c}\:. \\ $$

Commented by arvinddayama01@gmail.com last updated on 06/Oct/18

thanks

$$\mathrm{thanks} \\ $$

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