Question Number 60413 by tanmay last updated on 20/May/19
Commented by Meritguide1234 last updated on 21/May/19
$$\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{same}\:\mathrm{question}\:\mathrm{and}\:\mathrm{solition}\:\mathrm{from}\:\mathrm{goiit}\:\mathrm{page}\:\mathrm{by}\:\mathrm{Sourav}\:\mathrm{De} \\ $$
Commented by MJS last updated on 21/May/19
$$\mathrm{why}\:\mathrm{not}? \\ $$
Commented by tanmay last updated on 21/May/19
$${we}\:{are}\:{all}\:{students}…{we}\:{are}\:{not}\:\:{Tomas}\:{hardy}\: \\ $$$${or}\:{Ramanujam}…{we}\:{learn}\:{from}\:\:{each}\:{other} \\ $$$${what}\:{do}\left[{you}\:{want}\:{to}\:{say}..\:{i}\:{just}\:{shared}\:{the}\:{solution}\right. \\ $$$${to}\:{others}…{i}\:{am}\:{not}\:{in}\:{the}\:{side}\:{of}\:{others} \\ $$$${who}\:{think}\:{that}\:{they}\:{know}\:{every}\:{thing}…{i}\:{am}\:{still} \\ $$$${a}\:{student}…{i}\:{have}\:{miles}\:{to}\:{go}\:.. \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
$${this}\:{platform}\:{is}\:{open}\:{to}\:{all}\:{kind}\:{of}\:{exercices}\:{we}\:{do}\:{not}\:{distinguich}\: \\ $$$${that}\:{and}\:\:{that}\:\:{thanks}\:{to}\:{sir}\:{Tinkutara}…. \\ $$$$ \\ $$
Answered by tanmay last updated on 21/May/19
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}+\mathrm{1}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}\left(\mathrm{1}−{x}\right)−\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}−\mathrm{2}{x}−\mathrm{2}+\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$${now}\:\:{N}_{{r}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{{x}}{{x}+\mathrm{1}}+\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}{\mathrm{1}−\frac{{x}\left(\mathrm{1}−{x}\right)}{\left({x}+\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{2}−{x}−{x}+{x}^{\mathrm{2}} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)={D}_{{r}} \\ $$$${so}\:\frac{{N}_{{r}} }{{D}_{{r}} }=\mathrm{1} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}×{dx} \\ $$$$\mathrm{2}{I}=\mathrm{1} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by Meritguide1234 last updated on 21/May/19
