0-ln-cosh-x-cosh-x-dx-0-pi-2-log-1-sin-x-dx-

Question Number 141419 by mnjuly1970 last updated on 18/May/21

𝛗:=∫_0 ^( ∞) ((ln(cosh(x)))/(cosh(x)))dx=^? ∫_0 ^( (π/2)) log((1/(sin(x))))dx

$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({cosh}\left({x}\right)\right)}{{cosh}\left({x}\right)}{dx}\overset{?} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {log}\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\right){dx} \\ $$

Answered by mindispower last updated on 18/May/21

t=ch(t) =∫_1 ^∞ ((ln(t))/t).(dt/( (√(t^2 −1)))) =∫_1 ^∞ ((ln(t))/(t^2 (√(1−(1/t^2 )))))dt let (1/t)=sin(y) =∫_(π/2) ^0 ((ln((1/(sin(y)))))/( (√(1−sin^2 (y))))).−cos(y)dy =∫_0 ^(π/2) ln((1/(sin(y))))dy

$${t}={ch}\left({t}\right) \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}}.\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}{dt} \\ $$$${let}\:\frac{\mathrm{1}}{{t}}={sin}\left({y}\right) \\ $$$$=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \frac{{ln}\left(\frac{\mathrm{1}}{{sin}\left({y}\right)}\right)}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({y}\right)}}.−{cos}\left({y}\right){dy} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{{sin}\left({y}\right)}\right){dy} \\ $$

Commented by mindispower last updated on 18/May/21

$${pleasur} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 18/May/21

$${thanks}\:{alot}\:{sir}\:{power}.. \\ $$

Answered by mnjuly1970 last updated on 18/May/21

:=(1/2)∫_0 ^( ∞) ((cosh(x)ln(1+sinh^2 (x)))/(cosh^2 (x)))dx :=^(sinh(x)=t) (1/2)∫_0 ^( ∞) ((ln(1+t^2 ))/(1+t^2 ))dt :=^(t=tan(y)) (1/2)∫_0 ^( (π/2)) ((ln(1+tan^2 (y)))/1)dy :=(1/2)∫_0 ^( (π/2)) ln((1/(cos(y))))^2 dy :=∫_0 ^( (π/2)) −ln(cos(y))dy=∫_0 ^( (π/2)) −ln(sin(y))dy ............ 𝛗:=∫_0 ^( (π/2)) ln((1/(sin(x))))dx=(π/2)ln(2).......... ..........m.n.july.1970..........

$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{cosh}\left({x}\right){ln}\left(\mathrm{1}+{sinh}^{\mathrm{2}} \left({x}\right)\right)}{{cosh}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::\overset{{sinh}\left({x}\right)={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\::\overset{{t}={tan}\left({y}\right)} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({y}\right)\right)}{\mathrm{1}}{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{{cos}\left({y}\right)}\right)^{\mathrm{2}} {dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\::=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} −{ln}\left({cos}\left({y}\right)\right){dy}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} −{ln}\left({sin}\left({y}\right)\right){dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:…………\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\right){dx}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)………. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{m}.{n}.{july}.\mathrm{1970}………. \\ $$

Facebook Tweet Pin

0-ln-cosh-x-cosh-x-dx-0-pi-2-log-1-sin-x-dx-

Leave a Reply Cancel reply