Question Number 97173 by 675480065 last updated on 06/Jun/20
Answered by Sourav mridha last updated on 06/Jun/20
![I=(2/𝛑)∫_(−(π/4)) ^((+π)/4) (dx/((1+e^(sinx) )(2−cos2x)))...(i) and also I=(2/𝛑)∫_(+(π/4)) ^(−(𝛑/4)) ((d(−x))/((1+e^(sin(−x)) )(2−cos2(−x)))) =(2/𝛑)∫_(−(𝛑/4)) ^(+(𝛑/4)) (e^(sin(x)) /((1+e^(sin(x)) )(2−cos2x)))dx.....(ii) now (i)+(ii) we get, 2I=(2/𝛑)∫_(−(𝛑/4)) ^(+(𝛑/4)) (dx/(2sin^2 (x)+1)) [even f^n ]so I=−(2/𝛑)∫_0 ^(+(𝛑/4)) ((d(cotx))/(((√3))^2 +(cot(x))^2 )) =−(2/𝛑).(1/( (√3)))[tan^(−1) (((cotx)/( (√3))))]_0 ^(+(𝛑/4)) =−(2/𝛑).(1/( (√3)))[(𝛑/6)−(𝛑/2)]=(2/(3(√3))) so now I^2 =(4/(27))⇒27I^2 =4](https://www.tinkutara.com/question/Q97177.png)
$$\:\boldsymbol{{I}}=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{+\pi}{\mathrm{4}}} \frac{\boldsymbol{{dx}}}{\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sinx}}} \right)\left(\mathrm{2}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)}…\left(\boldsymbol{{i}}\right) \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{also}}\:\boldsymbol{{I}}=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{+\frac{\pi}{\mathrm{4}}} ^{−\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\mathrm{d}\left(−\mathrm{x}\right)}{\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sin}}\left(−\boldsymbol{{x}}\right)} \right)\left(\mathrm{2}−\boldsymbol{{cos}}\mathrm{2}\left(−\boldsymbol{{x}}\right)\right)} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{−\frac{\boldsymbol{\pi}}{\mathrm{4}}} ^{+\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\boldsymbol{{e}}^{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)} }{\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)} \right)\left(\mathrm{2}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)}\boldsymbol{{dx}}…..\left(\boldsymbol{{ii}}\right) \\ $$$$\boldsymbol{{now}}\:\left(\boldsymbol{{i}}\right)+\left(\boldsymbol{{ii}}\right)\:\boldsymbol{{we}}\:\boldsymbol{{get}}, \\ $$$$\mathrm{2}\boldsymbol{{I}}=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{−\frac{\boldsymbol{\pi}}{\mathrm{4}}} ^{+\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\boldsymbol{{dx}}}{\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)+\mathrm{1}}\:\:\left[\boldsymbol{{even}}\:\boldsymbol{{f}}^{\boldsymbol{{n}}} \right]\boldsymbol{{so}} \\ $$$$\boldsymbol{{I}}=−\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{\mathrm{0}} ^{+\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\boldsymbol{{d}}\left(\boldsymbol{{cotx}}\right)}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\boldsymbol{{cot}}\left(\boldsymbol{{x}}\right)\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=−\frac{\mathrm{2}}{\boldsymbol{\pi}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{cotx}}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{+\frac{\boldsymbol{\pi}}{\mathrm{4}}} \: \\ $$$$\:\:\:\:\:=−\frac{\mathrm{2}}{\boldsymbol{\pi}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\frac{\boldsymbol{\pi}}{\mathrm{6}}−\frac{\boldsymbol{\pi}}{\mathrm{2}}\right]=\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{now}}\:\boldsymbol{{I}}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{27}}\Rightarrow\mathrm{27}\boldsymbol{{I}}^{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$
Commented by 675480065 last updated on 06/Jun/20
$$\mathrm{i}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{from}\:\left(\mathrm{i}\right)+\left(\mathrm{ik}\right) \\ $$
Commented by Sourav mridha last updated on 07/Jun/20
$$\boldsymbol{{try}}\:\boldsymbol{{to}}\:\boldsymbol{{do}}\:\boldsymbol{{step}}\:\boldsymbol{{by}}\:\boldsymbol{{step}}\:−− \\ $$$$\boldsymbol{{at}}\:\boldsymbol{{first}}\:\boldsymbol{{add}}\:\left(\boldsymbol{{i}}\right)+\left(\boldsymbol{{ii}}\right) \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{you}}\:\boldsymbol{{will}}\:\boldsymbol{{see}}\:\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)} \right)\:\boldsymbol{{is}}\:\boldsymbol{{vanish}}. \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{its}}\:\boldsymbol{{remain}}\:\left(\mathrm{2}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{at}}\:\boldsymbol{{denomenator}}. \\ $$$$\boldsymbol{{this}}\:=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)+\mathrm{1}}=\frac{\boldsymbol{{cosec}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)}{\mathrm{3}+\boldsymbol{{cot}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \\ $$$$\boldsymbol{{no}}\mathrm{w}\:\boldsymbol{{goes}}\:\boldsymbol{{on}}\:\boldsymbol{{and}}\:\boldsymbol{{cheack}}\:\boldsymbol{{with}}\:\boldsymbol{{my}} \\ $$$$\boldsymbol{{sol}}^{\boldsymbol{{n}}} . \\ $$$$ \\ $$