Question-64687

Question Number 64687 by aliesam last updated on 20/Jul/19

Commented by mathmax by abdo last updated on 21/Jul/19

let A =∫ (e^((−cos(2x))/2) /(sin^2 x)) dx ⇒ A =∫ (e^(−((cos(2x))/2)) /((1−cos(2x))/2)) dx = 2 ∫ (e^(−((cos(2x))/2)) /(1−cos(2x)))dx changement cos(2x) =t give 2x=argch(t) =ln(t+(√(t^2 −1))) ⇒x =(1/2)ln(t+(√(t^2 −1))) ⇒dx =(1/2) ((1+((2t)/(2(√(t^2 −1)))))/(t+(√(t^2 −1)))) =(1/(2(√(t^2 −1)))) ⇒ A =2∫ (e^(−(t/2)) /(1−t)) (dt/(2(√(t^2 −1)))) = −∫ (e^(−(t/2)) /((t−1)(√(t^2 −1))))dt cha7gement t =ch(u)give A =−∫ (e^(−((chu)/2)) /((chu−1)shu)) sh(u)du = ∫ (e^(−((chu)/2)) /(1−ch(u)))du =∫ (e^(−((chu)/2)) /(1−((e^u +e^(−u) )/2))) du =∫ ((2 e^(−((chu)/2)) )/(2−e^u −e^(−u) )) du =∫ ((2e^(−u) e^(−((chu)/2)) )/(2e^(−u) −1−e^(−2u) ))du...be continued....

$${let}\:{A}\:=\int\:\:\:\frac{{e}^{\frac{−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} }{{sin}^{\mathrm{2}} {x}}\:{dx}\:\Rightarrow\:{A}\:=\int\:\:\:\frac{{e}^{−\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} }{\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}\:{dx} \\ $$$$=\:\mathrm{2}\:\int\:\:\:\frac{{e}^{−\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} }{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{dx}\:\:\:{changement}\:{cos}\left(\mathrm{2}{x}\right)\:={t}\:{give}\:\mathrm{2}{x}={argch}\left({t}\right) \\ $$$$={ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow{x}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:\:\:\Rightarrow{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}}{{t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:\:\Rightarrow\:{A}\:=\mathrm{2}\int\:\:\frac{{e}^{−\frac{{t}}{\mathrm{2}}} }{\mathrm{1}−{t}}\:\:\frac{{dt}}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\:−\int\:\:\:\:\frac{{e}^{−\frac{{t}}{\mathrm{2}}} }{\left({t}−\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt}\:\:\:\:{cha}\mathrm{7}{gement}\:\:\:{t}\:={ch}\left({u}\right){give} \\ $$$${A}\:=−\int\:\:\:\frac{{e}^{−\frac{{chu}}{\mathrm{2}}} }{\left({chu}−\mathrm{1}\right){shu}}\:{sh}\left({u}\right){du}\:=\:\int\:\:\:\frac{{e}^{−\frac{{chu}}{\mathrm{2}}} }{\mathrm{1}−{ch}\left({u}\right)}{du} \\ $$$$=\int\:\:\frac{{e}^{−\frac{{chu}}{\mathrm{2}}} }{\mathrm{1}−\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}\:{du}\:=\int\:\:\:\frac{\mathrm{2}\:{e}^{−\frac{{chu}}{\mathrm{2}}} }{\mathrm{2}−{e}^{{u}} −{e}^{−{u}} }\:{du} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{e}^{−{u}} \:{e}^{−\frac{{chu}}{\mathrm{2}}} }{\mathrm{2}{e}^{−{u}} −\mathrm{1}−{e}^{−\mathrm{2}{u}} }{du}…{be}\:{continued}…. \\ $$$$ \\ $$

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