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Question Number 116216 by bemath last updated on 02/Oct/20

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

Answered by Bird last updated on 02/Oct/20

$$letf\left(a\right)={\int }_0^{\pi }\frac{ln(1+acosx)}{cosx}dxwith-1<a<1$$$$$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\pi }\frac{dx}{1+acosx}$$$${=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{\mathrm{\infty }}\frac{2dt}{(1+t^2)(1+a.\frac{1-t^2}{1+t^2})}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2+a-{at}^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(1-a)t^2+1+a}$$$$=\frac{2}{(1-a)}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+\frac{1+a}{1-a}}$$$${=}_{t=\sqrt{\frac{1+a}{1+a}}u}\frac{2}{1-a}.\frac{1-a}{1+1}{\int }_0^{\mathrm{\infty }}\frac{1}{u^2+1}.\frac{\sqrt{1+a}}{1-a}u$$$$=\frac{2}{\sqrt{1-a^2}}.\frac{\pi }{2}=\frac{\pi }{\sqrt{1-a^2}}\Rightarrow$$$$f\left(a\right)=\int \frac{\pi }{\sqrt{1-a^2}}da+c$$$$=\pi arcsina+c$$$$f\left(0\right)=0=c\Rightarrow f\left(a\right)=\pi arcsin\left(a\right)$$$$and{\int }_0^{\pi }\frac{ln(1+\frac{1}{2}cosx)}{cosx}dx$$$$=f\left(\frac{1}{2}\right)=\pi arcsin\left(\frac{1}{2}\right)=\frac{{\pi }^2}{6}$$$$$$

Commented by bemath last updated on 02/Oct/20

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

Commented by mnjuly1970 last updated on 02/Oct/20

$$verynicesolution.thankyou$$$$mrbird..$$

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

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

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