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Question Number 171198 by infinityaction last updated on 09/Jun/22
evaluate ∫_0 ^( π) log (a+cos x)dx
$$ \\ $$$$\:\:{evaluate} \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} \mathrm{log}\:\left({a}+\mathrm{cos}\:{x}\right){dx} \\ $$
Answered by aleks041103 last updated on 09/Jun/22
I(a)=∫_0 ^π ln(a+cosx)dx I ′(a)=∫_0 ^π (dx/(a+cos(x))) t=tg(x/2) ⇒cosx=1−2sin^2 (x/2)= =1−(2/(csc^2 (x/2)))=1−(2/(1+ctg^2 (x/2)))= =1−(2/(1+(1/(tg^2 (x/2)))))=1−(2/(1+(1/t^2 )))= =1−((2t^2 )/(1+t^2 ))=((1−t^2 )/(1+t^2 )) x=2arctg(t)⇒dx=((2dt)/(1+t^2 )) x=0→t=0 x=π→t→∞ ⇒I ′(a)=∫_0 ^∞ ((2dt)/(a(1+t^2 )+1−t^2 ))= =2∫_0 ^∞ (dt/((a−1)t^2 +a+1))= =(2/(a+1))∫_0 ^∞ (dt/( ((√((a−1)/(a+1)))t)^2 +1))= =(2/(a+1))(√((a+1)/(a−1)))∫_0 ^∞ ((d((√((a−1)/(a+1)))t))/(((√((a−1)/(a+1)))t)^2 +1))= =(2/( (√(a^2 −1)))) (π/2)=(π/( (√(a^2 −1))))=I ′(a) ⇒I(a)=I(1)+π∫_1 ^a (da/( (√(a^2 −1)))) a=cosh(x) da=sinh(x)dx ⇒∫(da/( (√(a^2 −1))))=∫((sinh(x)dx)/( (√(cosh^2 x−1))))=x=arccosh(a) ⇒I(a)=I(1)+π(arccosh(a)−arccosh(1)) I(a)=I(1)+π arccosh(a) I(1)=∫_0 ^π ln(1+cos(x))dx ln(1+cos(x))=ln(2cos^2 (x/2))= =ln(2)+2ln(cos(x/2)) ⇒I(1)=πln(2)+4∫_0 ^(π/2) ln(cos(x))dx ∫_0 ^(π/2) ln(cosx)dx=∫_0 ^(π/2) ln(sinx)dx=i 2i=∫_0 ^(π/2) ln(sinxcosx)dx=∫_0 ^(π/2) ln((1/2)sin(2x))dx= =−((πln(2))/2)+∫_0 ^(π/2) ln(sin(2x))dx= =−((πln(2))/2)+(1/2)∫_0 ^π ln(sin(u))du= =−((πln(2))/2)+i=2i ⇒i=−((πln(2))/2) ⇒I(1)=πln(2)+4(−((πln(2))/2))=−πln(2) ⇒I(a)=π(arccosh(a)−ln(2)) cosh(x)=((e^x +e^(−x) )/2)=((t+(1/t))/2)=k ⇒2kt=t^2 +1 ⇒t^2 −2kt+1=0 ⇒t=e^x =((2k±(√(4k^2 −4)))/2)=k±(√(k^2 −1)) ⇒arccosh(x)=ln(x+(√(x^2 −1))) ∫_0 ^( π) ln(a+cos(x))dx=π ln(((a+(√(a^2 −1)))/2))
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\pi} {ln}\left({a}+{cosx}\right){dx} \\ $$$${I}\:'\left({a}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{a}+{cos}\left({x}\right)} \\ $$$${t}={tg}\left({x}/\mathrm{2}\right) \\ $$$$\Rightarrow{cosx}=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({x}/\mathrm{2}\right)= \\ $$$$=\mathrm{1}−\frac{\mathrm{2}}{{csc}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+{ctg}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}= \\ $$$$=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{{tg}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}= \\ $$$$=\mathrm{1}−\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${x}=\mathrm{2}{arctg}\left({t}\right)\Rightarrow{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${x}=\mathrm{0}\rightarrow{t}=\mathrm{0} \\ $$$${x}=\pi\rightarrow{t}\rightarrow\infty \\ $$$$\Rightarrow{I}\:'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{dt}}{{a}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+\mathrm{1}−{t}^{\mathrm{2}} }= \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({a}−\mathrm{1}\right){t}^{\mathrm{2}} +{a}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}}{{a}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\:\left(\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\right)^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}}{{a}+\mathrm{1}}\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}\int_{\mathrm{0}} ^{\infty} \frac{{d}\left(\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\right)}{\left(\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{t}\right)^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\frac{\pi}{\mathrm{2}}=\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}={I}\:'\left({a}\right) \\ $$$$\Rightarrow{I}\left({a}\right)={I}\left(\mathrm{1}\right)+\pi\int_{\mathrm{1}} ^{{a}} \frac{{da}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${a}={cosh}\left({x}\right) \\ $$$${da}={sinh}\left({x}\right){dx} \\ $$$$\Rightarrow\int\frac{{da}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}=\int\frac{{sinh}\left({x}\right){dx}}{\:\sqrt{{cosh}^{\mathrm{2}} {x}−\mathrm{1}}}={x}={arccosh}\left({a}\right) \\ $$$$\Rightarrow{I}\left({a}\right)={I}\left(\mathrm{1}\right)+\pi\left({arccosh}\left({a}\right)−{arccosh}\left(\mathrm{1}\right)\right) \\ $$$${I}\left({a}\right)={I}\left(\mathrm{1}\right)+\pi\:{arccosh}\left({a}\right) \\ $$$$ \\ $$$${I}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cos}\left({x}\right)\right){dx} \\ $$$${ln}\left(\mathrm{1}+{cos}\left({x}\right)\right)={ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({x}/\mathrm{2}\right)\right)= \\ $$$$={ln}\left(\mathrm{2}\right)+\mathrm{2}{ln}\left({cos}\left({x}/\mathrm{2}\right)\right) \\ $$$$\Rightarrow{I}\left(\mathrm{1}\right)=\pi{ln}\left(\mathrm{2}\right)+\mathrm{4}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({cos}\left({x}\right)\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({cosx}\right){dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sinx}\right){dx}={i} \\ $$$$\mathrm{2}{i}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sinxcosx}\right){dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right){dx}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\left({u}\right)\right){du}= \\ $$$$=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+{i}=\mathrm{2}{i} \\ $$$$\Rightarrow{i}=−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{I}\left(\mathrm{1}\right)=\pi{ln}\left(\mathrm{2}\right)+\mathrm{4}\left(−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\right)=−\pi{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow{I}\left({a}\right)=\pi\left({arccosh}\left({a}\right)−{ln}\left(\mathrm{2}\right)\right) \\ $$$${cosh}\left({x}\right)=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\frac{{t}+\frac{\mathrm{1}}{{t}}}{\mathrm{2}}={k} \\ $$$$\Rightarrow\mathrm{2}{kt}={t}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{kt}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}={e}^{{x}} =\frac{\mathrm{2}{k}\pm\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}={k}\pm\sqrt{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{arccosh}\left({x}\right)={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\pi} {ln}\left({a}+{cos}\left({x}\right)\right){dx}=\pi\:{ln}\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}}\right) \\ $$
Commented by infinityaction last updated on 10/Jun/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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