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Question Number 2212 by Yozzi last updated on 09/Nov/15
∫(dt/((1−kt)(√(1−t^2 ))))=? 0<k<1
$$\int\frac{{dt}}{\left(\mathrm{1}−{kt}\right)\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=?\:\:\mathrm{0}<{k}<\mathrm{1} \\ $$
Commented by 123456 last updated on 09/Nov/15
((ln ((√(k^2 −1))(√(1−x^2 ))+k−x)−ln (1−kx))/( (√(k^2 −1)))) c.m
$$\frac{\mathrm{ln}\:\left(\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{k}−{x}\right)−\mathrm{ln}\:\left(\mathrm{1}−{kx}\right)}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\mathrm{c}.\mathrm{m} \\ $$
Commented by prakash jain last updated on 09/Nov/15
t=sin u dt=cos udu ∫(du/(1−ksin u)) integrate with tan(u/2)=v sin u=((2v)/(1+v^2 )) cos u=((1−v^2 )/(1+v^2 )) (1/2)sec^2 (u/2)du=dv du=((2dv)/((1+v^2 ))) ∫(1/((1−k((2v)/(1+v^2 )))))∙((2dv)/((1+v^2 )))=∫((2dv)/(1+v^2 −2kv)) to be continued in answer.
$${t}=\mathrm{sin}\:{u} \\ $$$${dt}=\mathrm{cos}\:{udu} \\ $$$$\int\frac{{du}}{\mathrm{1}−{k}\mathrm{sin}\:{u}} \\ $$$$\mathrm{integrate}\:\mathrm{with}\:\mathrm{tan}\frac{{u}}{\mathrm{2}}={v} \\ $$$$\mathrm{sin}\:{u}=\frac{\mathrm{2}{v}}{\mathrm{1}+{v}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:{u}=\frac{\mathrm{1}−{v}^{\mathrm{2}} }{\mathrm{1}+{v}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}{du}={dv} \\ $$$${du}=\frac{\mathrm{2}{dv}}{\left(\mathrm{1}+{v}^{\mathrm{2}} \right)} \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}−{k}\frac{\mathrm{2}{v}}{\mathrm{1}+{v}^{\mathrm{2}} }\right)}\centerdot\frac{\mathrm{2}{dv}}{\left(\mathrm{1}+{v}^{\mathrm{2}} \right)}=\int\frac{\mathrm{2}{dv}}{\mathrm{1}+{v}^{\mathrm{2}} −\mathrm{2}{kv}} \\ $$$${to}\:{be}\:{continued}\:{in}\:{answer}. \\ $$
Answered by Filup last updated on 09/Nov/15
k=0 ∫(1/((1−0t)(√(1−t^2 ))))dt=∫(1/( (√(1−t^2 )))) t=sin θ dt=cos θdθ =∫((cos θ)/( (√(1−sin^2 θ))))dθ =∫((cos θ)/(cos θ))dθ =∫dθ=θ+c c=constant =sin^(−1) (t)+c −1≤t≤1 (1) k=1 ∫(1/((1−t)(√(1−t^2 ))))dt t=sin θ dt=cos θdθ =∫((cos θ)/((1−sin θ)(√(1−sin^2 θ))))dθ =∫((cos θ)/((1−sin θ)cos θ))dθ =∫(1/(1−sin θ))dθ (2) (continue) sin^(−1) (t)<∫(1/((1−kt)(√(1−t^2 ))))dt<∫(1/(1−sin θ))dθ
$${k}=\mathrm{0} \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{0}{t}\right)\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$${t}=\mathrm{sin}\:\theta\:\:\:\:\:\:\:\:{dt}=\mathrm{cos}\:\theta{d}\theta \\ $$$$=\int\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}}{d}\theta \\ $$$$=\int\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\theta}{d}\theta \\ $$$$=\int{d}\theta=\theta+{c}\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{constant} \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{t}\right)+{c}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\leqslant{t}\leqslant\mathrm{1}\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${k}=\mathrm{1} \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$${t}=\mathrm{sin}\:\theta\:\:\:\:\:\:\:\:\:{dt}=\mathrm{cos}\:\theta{d}\theta \\ $$$$=\int\frac{\mathrm{cos}\:\theta}{\left(\mathrm{1}−\mathrm{sin}\:\theta\right)\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}}{d}\theta \\ $$$$=\int\frac{\mathrm{cos}\:\theta}{\left(\mathrm{1}−\mathrm{sin}\:\theta\right)\mathrm{cos}\:\theta}{d}\theta \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\theta}{d}\theta\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{continue}\right) \\ $$$$ \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left({t}\right)<\int\frac{\mathrm{1}}{\left(\mathrm{1}−{kt}\right)\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}<\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\theta}{d}\theta \\ $$$$ \\ $$
Commented by Filup last updated on 09/Nov/15
I′m not sure how to finish this problem so i hope this is partially correct and on the right track
$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{this}\:\mathrm{problem} \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{is}\:\mathrm{partially}\:\mathrm{correct}\:\mathrm{and}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{right}\:\mathrm{track} \\ $$
Answered by prakash jain last updated on 09/Nov/15
∫((2dv)/(1+v^2 −2kv))=∫((2dv)/((v−k)^2 +((√(1−k^2 )))^2 )) =(2/( (√(1−k^2 ))))tan^(−1) ((v−k)/( (√(1−k^2 )))) v=tan (u/2),u=sin^(−1) x
$$\int\frac{\mathrm{2}{dv}}{\mathrm{1}+{v}^{\mathrm{2}} −\mathrm{2}{kv}}=\int\frac{\mathrm{2}{dv}}{\left({v}−{k}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}\mathrm{tan}^{−\mathrm{1}} \frac{{v}−{k}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }} \\ $$$${v}=\mathrm{tan}\:\frac{{u}}{\mathrm{2}},{u}=\mathrm{sin}^{−\mathrm{1}} {x} \\ $$

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