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 116216 by bemath last updated on 02/Oct/20

∫_0 ^π ((ln (1+(1/2)cos x))/(cos x)) dx ?

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Answered by Bird last updated on 02/Oct/20

let f(a) =∫_0 ^π ((ln(1+acosx))/(cosx))dx with−1<a<1 f^′ (a) =∫_0 ^π (dx/(1+acosx)) =_(tan((x/2))=t) ∫_0 ^∞ ((2dt)/((1+t^2 )(1+a.((1−t^2 )/(1+t^2 ))))) =2∫_0 ^∞ (dt/(1+t^2 +a−at^2 )) =2∫_0 ^∞ (dt/((1−a)t^2 +1+a)) =(2/((1−a)))∫_0 ^∞ (dt/(t^2 +((1+a)/(1−a)))) =_(t=(√((1+a)/(1+a)))u) (2/(1−a)).((1−a)/(1+1))∫_0 ^∞ (1/(u^2 +1)).((√(1+a))/(1−a))u =(2/( (√(1−a^2 )))).(π/2) =(π/( (√(1−a^2 )))) ⇒ f(a) =∫ (π/( (√(1−a^2 ))))da +c =π arcsina +c f(0) =0=c ⇒f(a) =π arcsin(a) and ∫_0 ^π ((ln(1+(1/2)cosx))/(cosx))dx =f((1/2)) =π arcsin((1/2))=(π^2 /6)

$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{ln}\left(\mathrm{1}+{acosx}\right)}{{cosx}}{dx}\:{with}−\mathrm{1}<{a}<\mathrm{1} \\ $$$$ \\ $$$${f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{acosx}} \\ $$$$=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}.\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} +{a}−{at}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}−{a}\right){t}^{\mathrm{2}} \:+\mathrm{1}+{a}} \\ $$$$=\frac{\mathrm{2}}{\left(\mathrm{1}−{a}\right)}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}} \\ $$$$=_{{t}=\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}+{a}}}{u}} \:\:\:\frac{\mathrm{2}}{\mathrm{1}−{a}}.\frac{\mathrm{1}−{a}}{\mathrm{1}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}.\frac{\sqrt{\mathrm{1}+{a}}}{\mathrm{1}−{a}}{u} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}.\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\int\:\:\frac{\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{da}\:+{c} \\ $$$$=\pi\:{arcsina}\:+{c} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}={c}\:\Rightarrow{f}\left({a}\right)\:=\pi\:{arcsin}\left({a}\right) \\ $$$${and}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{cosx}\right)}{{cosx}}{dx} \\ $$$$={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\pi\:{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by bemath last updated on 02/Oct/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mnjuly1970 last updated on 02/Oct/20

very nice solution .thank you mr bird..

$${very}\:{nice}\:{solution}\:.{thank}\:{you}\: \\ $$$${mr}\:{bird}.. \\ $$

Commented by mathmax by abdo last updated on 02/Oct/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com