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Question Number 60413 by tanmay last updated on 20/May/19

Commented by Meritguide1234 last updated on 21/May/19

why do you post same question and solition from goiit page by Sourav De

$$\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{same}\:\mathrm{question}\:\mathrm{and}\:\mathrm{solition}\:\mathrm{from}\:\mathrm{goiit}\:\mathrm{page}\:\mathrm{by}\:\mathrm{Sourav}\:\mathrm{De} \\ $$

Commented by MJS last updated on 21/May/19

why not?

$$\mathrm{why}\:\mathrm{not}? \\ $$

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[you want to say.. i just shared the solution 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 ..

$${we}\:{are}\:{all}\:{students}...{we}\:{are}\:{not}\:\:{Tomas}\:{hardy}\: \\ $$$${or}\:{Ramanujam}...{we}\:{learn}\:{from}\:\:{each}\:{other} \\ $$$${what}\:{do}\left[{you}\:{want}\:{to}\:{say}..\:{i}\:{just}\:{shared}\:{the}\:{solution}\right. \\ $$$${to}\:{others}...{i}\:{am}\:{not}\:{in}\:{the}\:{side}\:{of}\:{others} \\ $$$${who}\:{think}\:{that}\:{they}\:{know}\:{every}\:{thing}...{i}\:{am}\:{still} \\ $$$${a}\:{student}...{i}\:{have}\:{miles}\:{to}\:{go}\:.. \\ $$

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....

$${this}\:{platform}\:{is}\:{open}\:{to}\:{all}\:{kind}\:{of}\:{exercices}\:{we}\:{do}\:{not}\:{distinguich}\: \\ $$$${that}\:{and}\:\:{that}\:\:{thanks}\:{to}\:{sir}\:{Tinkutara}.... \\ $$$$ \\ $$

Answered by tanmay last updated on 21/May/19

∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx I=∫_0 ^1 ((tan^(−1) ((x/(x+1))))/(tan^(−1) (((1+2x−2x^2 )/2))))dx =∫_0 ^1 ((tan^(−1) (((1−x)/(1−x+1))))/(tan^(−1) (((1+2(1−x)−2(1−x)^2 )/2))))dx =∫_0 ^1 ((tan^(−1) (((1−x)/(2−x))))/(tan^(−1) (((1+2−2x−2+4x−2x^2 )/2))))dx =∫_0 ^1 ((tan^(−1) (((1−x)/(1+x))))/(tan^(−1) (((1+2x−2x^2 )/2))))dx 2I=∫_0 ^1 ((tan^(−1) ((x/(x+1)))+tan^(−1) (((1−x)/(2−x))))/(tan^(−1) (((1+2x−2x^2 )/2))))dx now N_r tan^(−1) ((x/(x+1)))+tan^(−1) (((1−x)/(2−x))) =tan^(−1) ((((x/(x+1))+((1−x)/(2−x)))/(1−((x(1−x))/((x+1)(2−x)))))) =tan^(−1) (((2x−x^2 +1−x^2 )/(2x−x^2 +2−x−x+x^2 ))) =tan^(−1) (((1+2x−2x^2 )/2))=D_r so (N_r /D_r )=1 2I=∫_0 ^1 1×dx 2I=1 I=(1/2)

$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}+\mathrm{1}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}\left(\mathrm{1}−{x}\right)−\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}−\mathrm{2}{x}−\mathrm{2}+\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right)}{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{dx} \\ $$$${now}\:\:{N}_{{r}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{1}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{{x}}{{x}+\mathrm{1}}+\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}{\mathrm{1}−\frac{{x}\left(\mathrm{1}−{x}\right)}{\left({x}+\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{2}−{x}−{x}+{x}^{\mathrm{2}} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\right)={D}_{{r}} \\ $$$${so}\:\frac{{N}_{{r}} }{{D}_{{r}} }=\mathrm{1} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}×{dx} \\ $$$$\mathrm{2}{I}=\mathrm{1} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Commented by Meritguide1234 last updated on 21/May/19

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