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Question Number 49806 by maxmathsup by imad last updated on 10/Dec/18

$$letf\left(x\right)={\int }_0^{\frac{\pi }{4}}ln(1-x^{2}cos\theta )d\theta with\mid x\mid <1$$ $$1)findaexplicitformoff\left(x\right)$$ $$2)calculate{\int }_0^{\frac{\pi }{4}}ln(1-\frac{1}{4}cos\theta )d\theta .$$

Commented byAbdo msup. last updated on 12/Dec/18

$$1)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{4}}\frac{-2xcos\theta }{1-x^{2}cos\theta }d\theta$$ $$=\frac{2}{x}{\int }_0^{\frac{\pi }{4}}\frac{1-x^{2}cos\theta -1}{1-x^{2}cos\theta }d\theta$$ $$=\frac{\pi }{2x}-\frac{2}{x}{\int }_0^{\frac{\pi }{4}}\frac{d\theta }{1-x^{2}cos\theta }letfind{\int }_0^{\frac{\pi }{4}}\frac{d\theta }{1-x^{2}cos\theta }$$ $$changementtan\left(\frac{\theta }{2}\right)=t\Rightarrow$$ $${\int }_0^{\frac{\pi }{4}}\frac{d\theta }{1-x^{2}cos\theta }={\int }_0^{\sqrt{2}-1}\frac{1}{1-x^2\frac{1-t^2}{1+t^2}}\frac{2t}{1+t^2}$$ $$={\int }_0^{\sqrt{2}-1}\frac{2dt}{1+t^2-x^2+x^2{t}^2}={\int }_0^{\sqrt{2}-1}\frac{2dt}{1-x^2+(1+x^2)t^2}$$ $$=\frac{2}{1-x^2}{\int }_0^{\sqrt{2}-1}\frac{dt}{1+\frac{1+x^2}{1-x^2}{t}^2}letsuppose\mid x\mid <1$$ $${=}_{\sqrt{\frac{1+x^2}{1-x^2}t}=u}\frac{2}{1-x^2}{\int }_0^{(\sqrt{2}-1)\sqrt{\frac{1+x^2}{1-x^2}}}\frac{1}{1+u^2}\sqrt{\frac{1-x^2}{1+x^2}}du$$ $$=\frac{2}{\sqrt{1-x^4}}arctan\left\{(\sqrt{2}-1)\sqrt{\frac{1+x^2}{1-x^2}}\right\}\Rightarrow$$ $$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{2x}-\frac{1}{x\sqrt{1-x^4}}arctan\left\{(\sqrt{2}-1)\sqrt{\frac{1+x^2}{1-x^2}}\right\}\Rightarrow$$ $$f\left(x\right)=\frac{\pi }{2}ln\mid x\mid -{\int }_1^x\frac{1}{t\sqrt{1-t^4}}arctan\left\{(\sqrt{2}-1)\sqrt{\frac{1+t^2}{1-t^2}}\right\}+\lambda$$

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