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Question Number 38899 by math khazana by abdo last updated on 01/Jul/18

$$find{\int }_0^{\pi }ln(2+cost)dtand{\int }_0^{\pi }ln(2-cost)dt$$

Commented by maxmathsup by imad last updated on 11/Jul/18

$$letI={\int }_0^{\pi }ln(2+cost)dt$$$$I=\pi ln\left(2\right)+{\int }_0^{\pi }ln(1+\frac{1}{2}cost)tletntroducetheparametricfunction$$$$f\left(x\right)={\int }_0^{\pi }ln(1+xcost)dtwehaveI=f\left(\frac{1}{2}\right)(wetake\mid x\mid <1)$$$$f^{{}^{\prime }}\left(x\right)={\int }_0^{\pi }\frac{cost}{1+xcost}dt=\frac{1}{x}{\int }_0^{\pi }\frac{1+xcost-1}{1+xcost}dt$$$$=\frac{\pi }{x}-\frac{1}{x}{\int }_0^{\pi }\frac{dt}{1+xcost}dtbutchangementtan\left(\frac{t}{2}\right)=ugive$$$${\int }_0^{\pi }\frac{dt}{1+xcost}={\int }_0^{\mathrm{\infty }}\frac{1}{1+x\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2du}{1+u^2+x(1-u^2)}={\int }_0^{\mathrm{\infty }}\frac{2du}{1+x+(1-x)u^2}$$$$=\frac{2}{1+x}{\int }_0^{\mathrm{\infty }}\frac{du}{1+\frac{1-x}{1+x}{u}^2}{=}_{\sqrt{\frac{1-x}{1+x}u}=\alpha }\frac{2}{1+x}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\alpha }^2}\sqrt{\frac{1+x}{1-x}}d\alpha$$$$=\frac{2}{\sqrt{1-x^2}}\frac{\pi }{2}=\frac{\pi }{\sqrt{1-x^2}}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{\pi }{x}-\frac{1}{x}\frac{\pi }{\sqrt{1-x^2}}\Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid -\pi {\int }_{}\frac{dx}{x\sqrt{1-x^2}}+c$$$$\int \frac{dx}{x\sqrt{1-x^2}}{=}_{x=sin\theta }\int \frac{cos\theta d\theta }{sin\theta cos\theta }=\int \frac{d\theta }{sin\theta }{=}_{tan\left(\frac{\theta }{2}\right)=u}\int \frac{1}{\frac{2u}{1+u^2}}\frac{2du}{1+u^2}$$$$=\int \frac{du}{u}=ln\mid u\mid =ln\mid tan\left(\frac{\theta }{2}\right)\mid =ln\mid tan\left(\frac{arcinx}{2}\right)\mid \Rightarrow$$$$f\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan\left(\frac{arcsinx}{2}\right)\mid +c$$$$wehavef\left(1\right)=c={\int }_0^{\pi }ln(1+cosx)dx={\int }_0^{\pi }ln\left(2{cos}^2\left(\frac{x}{2}\right)\right)dx$$$$=\pi ln\left(2\right)+2{\int }_0^{\pi }ln\left(cos\left(\frac{x}{2}\right)\right)dx$$$${=}_{\frac{x}{2}=t}\pi ln\left(2\right)+2{\int }_0^{\frac{\pi }{2}}ln\left(cost\right)\left(2dt\right)=\pi ln\left(2\right)+4(-\frac{\pi }{2}ln\left(2\right))$$$$=-\pi ln\left(2\right)\Rightarrow f\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan\left(\frac{arcsinx}{2}\right)\mid -\pi ln\left(2\right)$$$$I=f\left(\frac{1}{2}\right)=-2\pi ln\left(2\right)-\pi ln\mid tan\left(\frac{\pi }{12}\right)\mid$$$${cos}^2\left(\frac{\pi }{12}\right)=\frac{1+cos\left(\frac{\pi }{6}\right)}{2}=\frac{1+\frac{\sqrt{3}}{2}}{2}=\frac{2+\sqrt{3}}{4}\Rightarrow cos\left(\frac{\pi }{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}$$$${sin}^2\left(\frac{\pi }{12}\right)=\frac{1-cos\left(\frac{\pi }{6}\right)}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{2-\sqrt{3}}{4}\Rightarrow sin\left(\frac{\pi }{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{2}$$$$tan\left(\frac{\pi }{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\frac{2-\sqrt{3}}{\sqrt{4-3}}=2-\sqrt{3}.\Rightarrow$$$$I=-2\pi ln\left(2\right)-\pi ln(2-\sqrt{3})$$

Commented by maxmathsup by imad last updated on 11/Jul/18

$$letJ={\int }_0^{\pi }ln(2-cost)dt$$$$J=\pi ln\left(2\right)+{\int }_0^{\pi }ln(1-\frac{1}{2}cost)dt=\pi ln\left(2\right)+f(-\frac{1}{2})$$$$=\pi ln\left(2\right)+f\left(\frac{1}{2}\right)\left(fisodd\right)$$$$=\pi ln\left(2\right)-2\pi ln\left(2\right)-\pi ln(2-\sqrt{3})=-\pi ln\left(2\right)-\pi ln(2-\sqrt{3})$$$$$$

Commented by maxmathsup by imad last updated on 11/Jul/18

$$I=\pi ln\left(2\right)+f\left(\frac{1}{2}\right)=-\pi ln\left(2\right)-\pi ln(2-\sqrt{3}).$$

Commented by maxmathsup by imad last updated on 11/Jul/18

$$fiseven.$$

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