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Question Number 171198 by infinityaction last updated on 09/Jun/22

$$$$$$evaluate$$$${\int }_0^{\pi }\log (a+\cos x)dx$$

Answered by aleks041103 last updated on 09/Jun/22

$$I\left(a\right)={\int }_0^{\pi }ln(a+cosx)dx$$$$I{}^{\prime }\left(a\right)={\int }_0^{\pi }\frac{dx}{a+cos\left(x\right)}$$$$t=tg\left(x/2\right)$$$$\Rightarrow cosx=1-2{sin}^2\left(x/2\right)=$$$$=1-\frac{2}{{csc}^2\left(x/2\right)}=1-\frac{2}{1+{ctg}^2\left(x/2\right)}=$$$$=1-\frac{2}{1+\frac{1}{{tg}^2\left(x/2\right)}}=1-\frac{2}{1+\frac{1}{t^2}}=$$$$=1-\frac{2t^2}{1+t^2}=\frac{1-t^2}{1+t^2}$$$$x=2arctg\left(t\right)\Rightarrow dx=\frac{2dt}{1+t^2}$$$$x=0\to t=0$$$$x=\pi \to t\to \mathrm{\infty }$$$$\Rightarrow I{}^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{2dt}{a(1+t^2)+1-t^2}=$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(a-1)t^2+a+1}=$$$$=\frac{2}{a+1}{\int }_0^{\mathrm{\infty }}\frac{dt}{{\left(\sqrt{\frac{a-1}{a+1}}t\right)}^2+1}=$$$$=\frac{2}{a+1}\sqrt{\frac{a+1}{a-1}}{\int }_0^{\mathrm{\infty }}\frac{d\left(\sqrt{\frac{a-1}{a+1}}t\right)}{{\left(\sqrt{\frac{a-1}{a+1}}t\right)}^2+1}=$$$$=\frac{2}{\sqrt{a^2-1}}\frac{\pi }{2}=\frac{\pi }{\sqrt{a^2-1}}=I{}^{\prime }\left(a\right)$$$$\Rightarrow I\left(a\right)=I\left(1\right)+\pi {\int }_1^a\frac{da}{\sqrt{a^2-1}}$$$$a=cosh\left(x\right)$$$$da=sinh\left(x\right)dx$$$$\Rightarrow \int \frac{da}{\sqrt{a^2-1}}=\int \frac{sinh\left(x\right)dx}{\sqrt{{cosh}^{2}x-1}}=x=arccosh\left(a\right)$$$$\Rightarrow I\left(a\right)=I\left(1\right)+\pi (arccosh\left(a\right)-arccosh\left(1\right))$$$$I\left(a\right)=I\left(1\right)+\pi arccosh\left(a\right)$$$$$$$$I\left(1\right)={\int }_0^{\pi }ln(1+cos\left(x\right))dx$$$$ln(1+cos\left(x\right))=ln\left(2{cos}^2\left(x/2\right)\right)=$$$$=ln\left(2\right)+2ln\left(cos\left(x/2\right)\right)$$$$\Rightarrow I\left(1\right)=\pi ln\left(2\right)+4{\int }_0^{\pi /2}ln\left(cos\left(x\right)\right)dx$$$${\int }_0^{\pi /2}ln\left(cosx\right)dx={\int }_0^{\pi /2}ln\left(sinx\right)dx=i$$$$2i={\int }_0^{\pi /2}ln\left(sinxcosx\right)dx={\int }_0^{\pi /2}ln\left(\frac{1}{2}sin\left(2x\right)\right)dx=$$$$=-\frac{\pi ln\left(2\right)}{2}+{\int }_0^{\pi /2}ln\left(sin\left(2x\right)\right)dx=$$$$=-\frac{\pi ln\left(2\right)}{2}+\frac{1}{2}{\int }_0^{\pi }ln\left(sin\left(u\right)\right)du=$$$$=-\frac{\pi ln\left(2\right)}{2}+i=2i$$$$\Rightarrow i=-\frac{\pi ln\left(2\right)}{2}$$$$\Rightarrow I\left(1\right)=\pi ln\left(2\right)+4(-\frac{\pi ln\left(2\right)}{2})=-\pi ln\left(2\right)$$$$\Rightarrow I\left(a\right)=\pi (arccosh\left(a\right)-ln\left(2\right))$$$$cosh\left(x\right)=\frac{e^x+e^{-x}}{2}=\frac{t+\frac{1}{t}}{2}=k$$$$\Rightarrow 2kt=t^2+1$$$$\Rightarrow t^2-2kt+1=0$$$$\Rightarrow t=e^x=\frac{2k\pm \sqrt{4k^2-4}}{2}=k\pm \sqrt{k^2-1}$$$$\Rightarrow arccosh\left(x\right)=ln(x+\sqrt{x^2-1})$$$${\int }_0^{\pi }ln(a+cos\left(x\right))dx=\pi ln\left(\frac{a+\sqrt{a^2-1}}{2}\right)$$

Commented by infinityaction last updated on 10/Jun/22

$$thankyousir$$

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