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Question Number 84810 by M±th+et£s last updated on 16/Mar/20

$${\int }_0^{\pi }ln\left(\frac{1+bcos\left(x\right)}{1+asin\left(x\right)}\right)dx$$ $$-1<a<b<1$$

Commented bymathmax by abdo last updated on 16/Mar/20

$${\int }_0^{\pi }ln\left(\frac{1+bcosx}{1+asinx}\right)dx=g\left(b\right)-f\left(a\right)withg\left(b\right)={\int }_0^{\pi }ln(1+bcosx)dx$$ $$andf\left(a\right)={\int }_0^{\pi }ln(1+asinx)dxwehave{g}^{{}^{\prime }}\left(b\right)={\int }_0^{\pi }\frac{cosx}{1+bcosx}dx$$ $$=\frac{1}{b}{\int }_0^{\pi }\frac{bcosx+1-1}{1+bcosx}dx=\frac{\pi }{b}-\frac{1}{b}{\int }_0^{\pi }\frac{dx}{1+bcosx}$$ $${\int }_0^{\pi }\frac{dx}{1+bcosx}{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{\mathrm{\infty }}\frac{1}{1+b\frac{1-t^2}{1+t^2}}\times \frac{2dt}{1+t^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{1+t^2+b-{bt}^2}$$ $$={\int }_0^{\mathrm{\infty }}\frac{2dt}{(1-b)t^2+1+b}=\frac{2}{1-b}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+\frac{1+b}{1-b}}$$ $${=}_{t=\sqrt{\frac{1+b}{1-b}}u}\frac{2}{1-b}\times \frac{1-b}{1+b}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\sqrt{\frac{1+b}{1-b}}du$$ $$=\frac{2}{\sqrt{1-b^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-b^2}}\Rightarrow g^{{}^{\prime }}\left(b\right)=\frac{\pi }{b}-\frac{\pi }{b\sqrt{1-b^2}}\Rightarrow$$ $$g\left(b\right)=\pi ln\mid b\mid -\pi {\int }_b^1\frac{dx}{x\sqrt{1-x^2}}+K$$ $$g\left(1\right)=K$$ $$K={\int }_0^{\pi }ln(1+cosx)dx\Rightarrow g\left(b\right)=\pi ln\mid b\mid -\pi {\int }_b^1\frac{dx}{x\sqrt{1-x^2}}+{\int }_0^{\pi }ln(1+cosx)dx$$ $${\int }_b^1\frac{dx}{x\sqrt{1-x^2}}{=}_{x=sin\alpha }{\int }_{arcsinb}^{\frac{\pi }{2}}\frac{cos\alpha d\alpha }{sin\alpha .cos\alpha }={\int }_{arcsinb}^{\frac{\pi }{2}}\frac{d\alpha }{sin\alpha }$$ $${=}_{tan\left(\frac{\alpha }{2}\right)=z}{\int }_{tan\left(\frac{arcsinb}{2}\right)}^1\frac{2dz}{(1+z^2)\frac{2z}{1+z^2}}={\int }_{tan\left(\frac{arcsinb}{2}\right)}^1\frac{dz}{z}$$ $$={[ln\mid z\mid ]}_{tan\left(\frac{arcsinb}{2}\right)}^1=-ln\mid tan\left(\frac{arcsinb}{2}\right))\Rightarrow$$ $$g\left(b\right)=\pi ln\mid b\mid +\pi ln\mid tan\left(\frac{arcsinb}{2}\right)\mid +{\int }_0^{\pi }ln(1+cosx)dx$$ $${\int }_0^{\pi }ln(1+cosx)dx={\int }_0^{\pi }ln\left(2{cos}^2\left(\frac{x}{2}\right)\right)dx$$ $$=\pi ln\left(2\right)+2{\int }_0^{\pi }ln\left(cos\left(\frac{x}{2}\right)\right)dx(\frac{x}{2}=u)$$ $$=\pi ln\left(2\right)+2{\int }_0^{\frac{\pi }{2}}ln\left(cosu\right)\left(2du\right)$$ $$=\pi ln\left(2\right)+4(-\frac{\pi }{2}ln\left(2\right))=-\pi ln\left(2\right)\Rightarrow$$ $$g\left(b\right)=\pi (ln\mid b\mid -ln\left(2\right))+\pi ln\mid tan\left(\frac{arcsinb}{2}\right)\mid ...becontinued$$

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