Question Number 68043 by mhmd last updated on 03/Sep/19
$$\int_{\pi/\mathrm{2}} ^{\pi} {e}^{{cosx}} \sqrt{\mathrm{1}−{e}^{{cosx}} }\:{sinx}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 03/Sep/19
![let I =∫_(π/2) ^π e^(cosx) (√(1−e^(cosx) ))sinx dx vha7gement cosx=−t give I =∫_0 ^1 e^(−t) (√(1−e^(−t) ))(dt) =∫_0 ^1 e^(−t) (√(1−e^(−t) ))dt after we do the changement (√(1−e_ ^(−t) ))=u ⇒1−e^(−t) =u^2 ⇒e^(−t) =1−u^2 ⇒ −t =ln(1−u^2 ) ⇒t =−ln(1−u^2 ) ⇒ I =∫_0 ^(√(1−e^(−1) )) (1−u^2 )u ×((2u)/(1−u^2 ))du =2 ∫_0 ^(√(1−e^(−1) )) u^(2 ) du =(2/3)[u^3 ]_0 ^(√(1−e^(−1) )) =(2/3){(√(1−e^(−1) ))}^3 =(2/3)(1−e^(−1) )(√(1−(1/e))) =(2/3)(1−(1/e))((√(e−1))/( (√e))) =(2/3)(((e−1)/e))((√(e−1))/( (√e))) ⇒ I =((2(e−1)(√(e−1)))/(3e(√e))) .](https://www.tinkutara.com/question/Q68053.png)
$${let}\:{I}\:=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{e}^{{cosx}} \sqrt{\mathrm{1}−{e}^{{cosx}} }{sinx}\:{dx}\:\:{vha}\mathrm{7}{gement}\:\:{cosx}=−{t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} \sqrt{\mathrm{1}−{e}^{−{t}} }\left({dt}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} \sqrt{\mathrm{1}−{e}^{−{t}} }{dt}\:\:\:{after}\:{we}\:{do}\:{the} \\ $$$${changement}\:\sqrt{\mathrm{1}−{e}_{} ^{−{t}} }={u}\:\Rightarrow\mathrm{1}−{e}^{−{t}} \:={u}^{\mathrm{2}} \:\Rightarrow{e}^{−{t}} \:=\mathrm{1}−{u}^{\mathrm{2}} \:\Rightarrow \\ $$$$−{t}\:={ln}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\:\Rightarrow{t}\:=−{ln}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} \:\:\:\:\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}\:×\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} \:\:\:{u}^{\mathrm{2}\:} {du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[{u}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} \:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\left\{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }\right\}^{\mathrm{3}} \:=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{e}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right)\frac{\sqrt{{e}−\mathrm{1}}}{\:\sqrt{{e}}}\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{{e}−\mathrm{1}}{{e}}\right)\frac{\sqrt{{e}−\mathrm{1}}}{\:\sqrt{{e}}} \\ $$$$\:\Rightarrow\:{I}\:=\frac{\mathrm{2}\left({e}−\mathrm{1}\right)\sqrt{{e}−\mathrm{1}}}{\mathrm{3}{e}\sqrt{{e}}}\:. \\ $$
Answered by mr W last updated on 03/Sep/19
![=−∫_(π/2) ^π e^(cosx) (√(1−e^(cosx) )) d cos x =∫_(−1) ^0 e^t (√(1−e^t )) dt =∫_(−1) ^0 (√(1−e^t )) de^t =∫_(1/e) ^1 (√(1−u)) du =−∫_(1/e) ^1 (√(1−u)) d(1−u) =−(2/3)[(1−u)^(3/2) ]_(1/e) ^1 =(2/3)(1−(1/e))^(3/2)](https://www.tinkutara.com/question/Q68044.png)
$$=−\int_{\pi/\mathrm{2}} ^{\pi} {e}^{{cosx}} \sqrt{\mathrm{1}−{e}^{{cosx}} }\:{d}\:\mathrm{cos}\:{x} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} {e}^{{t}} \sqrt{\mathrm{1}−{e}^{{t}} }\:{dt} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \sqrt{\mathrm{1}−{e}^{{t}} }\:{de}^{{t}} \\ $$$$=\int_{\frac{\mathrm{1}}{{e}}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{u}}\:{du} \\ $$$$=−\int_{\frac{\mathrm{1}}{{e}}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{u}}\:{d}\left(\mathrm{1}−{u}\right) \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\frac{\mathrm{1}}{{e}}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$