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Question Number 38718 by maxmathsup by imad last updated on 28/Jun/18

$$1)findf\left(x\right)={\int }_0^{\pi }ln(2+xcos\theta )d\theta$$$$2)calculate{\int }_0^{\pi }ln(2+cos\theta )d\theta$$$$$$

Commented by math khazana by abdo last updated on 30/Jun/18

$$1)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\pi }\frac{cos\theta }{2+xcos\theta }d\theta changement$$$$tan\left(\frac{\theta }{2}\right)=t\Rightarrow f^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{\frac{1-t^2}{1+t^2}}{2+x\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$f^{{}^{\prime }}\left(x\right)=2{\int }_0^{\mathrm{\infty }}\frac{1-t^2}{(1+t^2)(2+2t^2+x-{xt}^2)}dt$$$$=2{\int }_0^{\mathrm{\infty }}\frac{1-t^2}{(1+t^2)\{(2-x)t^2+x+2\}}dtlet$$$$f^{{}^{\prime }}\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1-t^2}{(1+t^2)\{(2-x)t^2+x+2\}}dtlet$$$$\phi \left(z\right)=\frac{1-z^2}{(z^2+1)\{(2-x)z^2+x+2\}}ifx\ne 2$$$$\phi \left(z\right)=\frac{1-z^2}{(2-x)(z^2+1)\{z^2+\frac{2+x}{2-x}\}}$$$$case1\frac{2+x}{2-x}>0\Rightarrow$$$$\phi \left(z\right)=\frac{1-z^2}{(2-x)(z-i)(z+i)(z-i\sqrt{\frac{2+x}{2-x}})(z+i\sqrt{\frac{2+x}{2-x}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,i)+Res(\phi ,i\sqrt{\frac{2+x}{2-x}})\}$$$$Res(\phi ,i)=\frac{2}{(2-x)\left(2i\right)(-1+\frac{2+x}{2-x})}$$$$=\frac{-i(2-x)}{(2-x)(-2+x+2+x)}=\frac{-i}{2x}$$$$Res(\phi ,i\sqrt{\frac{2+x}{2-x}})=\frac{1+\frac{2+x}{2-x}}{(2-x)(-\frac{2+x}{2-x}+1)2i\sqrt{\frac{2+x}{2-x}}}$$$$=\frac{4}{{(2-x)}^2\left(\frac{-2-x+2-x}{2-x}\right)2i\sqrt{\frac{2+x}{2-x}}}$$$$=\frac{-2i}{(2-x)(-2x)\frac{\sqrt{2+x}}{\sqrt{2-x}}}=\frac{i}{\sqrt{4-x^2}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{-i}{2x}+\frac{i}{\sqrt{4-x^2}}\}$$$$=\frac{\pi }{x}+\frac{2\pi }{\sqrt{4-x^2}}=f^{{}^{\prime }}\left(x\right)\Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid +2\pi \int \frac{dx}{\sqrt{4-x^2}}+cchangement$$$$x=2ugive\int \frac{dx}{\sqrt{4-x^2}}=\int \frac{2du}{2\sqrt{1-u^2}}$$$$=arcsin\left(\frac{x}{2}\right)\Rightarrow f\left(x\right)=\pi ln\mid x\mid +2\pi arcsin\left(\frac{x}{2}\right)+c$$$$f\left(1\right)=2\pi \frac{\pi }{2}+c={\pi }^2+c\Rightarrow c=f\left(1\right)-{\pi }^2\Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid +2\pi arcsin\left(\frac{x}{2}\right)+f\left(1\right)-{\pi }^2.$$$$$$

Commented by math khazana by abdo last updated on 30/Jun/18

$$case2\frac{2+x}{2-x}<0wefollowthesamemethod..$$

Commented by math khazana by abdo last updated on 30/Jun/18

$$errorfromline15$$$$Res(\phi ,i\sqrt{\frac{2+x}{2-x}})=\frac{i}{x\sqrt{4-x^2}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{-i}{2x}+\frac{i}{x\sqrt{4-x^2}}\}$$$$=\frac{\pi }{x}-\frac{2\pi }{x\sqrt{4-x^2}}=f^{{}^{\prime }}\left(x\right)\Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid -2\pi \int \frac{dx}{x\sqrt{4-x^2}}.changement$$$$x=2sint\Rightarrow$$$$\int \frac{dx}{x\sqrt{4-x^2}}=\int \frac{2cost}{2sint2cost}dt$$$$=\frac{1}{2}\int \frac{dt}{sint}thenchang.tan\left(\frac{t}{2}\right)=u$$$$=\frac{1}{2}\int \frac{1}{\frac{2u}{1+u^2}}\frac{2du}{1+u^2}=\frac{1}{2}\int \frac{du}{u}=\frac{1}{2}ln\mid tan\left(\frac{t}{2}\right))$$$$=\frac{1}{2}ln\mid tan\left(\frac{1}{2}arcsin\left(\frac{x}{2}\right)\right)\mid \Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan(\frac{1}{2}arcsin\left(\frac{x}{2}\right)\mid +c$$$$f\left(1\right)=c\Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan(\frac{1}{2}arcsin\left(\frac{x}{2}\right)\mid +{\int }_0^{\pi }ln(2+cos\theta )d\theta$$

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