Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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....

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com