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Question Number 40661 by math khazana by abdo last updated on 25/Jul/18

$$1)findg\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1-x^2{cos}^2\theta )d\theta withxfromR$$$$2)findthevalueof{\int }_0^{\frac{\pi }{2}}ln(1-2{cos}^2\theta )d\theta and$$$$3)findthevalueof$$$$A\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}ln(1-{cos}^2\alpha {cos}^2\theta )d\theta$$

Commented by math khazana by abdo last updated on 04/Aug/18

$$1)wehaveg\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1+xcos\theta )d\theta$$$$+{\int }_0^{\frac{\pi }{2}}ln(1-xcos\theta )d\theta =h\left(x\right)+k\left(x\right)$$$$h\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1+xcos\theta )d\theta \Rightarrow$$$$h^{{}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{cos\theta }{1+xcos\theta }d\theta$$$$=\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{1+xcos\theta -1}{1+xcos\theta }d\theta =\frac{\pi }{2x}-\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+xcos\theta }$$$$but{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+xcos\theta }{=}_{tan\left(\frac{\theta }{2}\right)=t}{\int }_0^1\frac{1}{1+x\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$={\int }_0^1\frac{2dt}{1+t^2+x-{xt}^2}={\int }_0^1\frac{2dt}{1+x+(1-x)t^2}$$$$=\frac{2}{(1+x)}{\int }_0^1\frac{dt}{1+\frac{1-x}{1+x}{t}^2}if\mid x\mid <1wedothe$$$$changement\sqrt{\frac{1-x}{1+x}}t=u\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+xcos\theta }=\frac{2}{1+x}{\int }_0^{\sqrt{\frac{1-x}{1+x}}}\frac{1}{1+u^2}\sqrt{\frac{1+x}{1-x}}du$$$$=\frac{2}{\sqrt{1-x^2}}arctan\sqrt{\frac{1-x}{1+x}}\Rightarrow$$$$h\left(x\right)=\int \frac{2arctan\sqrt{\frac{1-x}{1+x}}}{\sqrt{1-x^2}}dx+c$$$$changementx=costgive$$$$h\left(x\right)=\int \frac{2arctan\left(tan\left(\frac{t}{2}\right)\right)}{sint}(-sint)dt+c$$$$=-\int tdt+c=-\frac{t^2}{2}+c=c-\frac{1}{2}{\left(arccosx\right)}^2$$$$h\left(0\right)=0=c-\frac{1}{2}\Rightarrow c=\frac{1}{2}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}ln(1+xcos\theta )d\theta =\frac{1}{2}-\frac{1}{2}{\left(arcosx\right)}^2$$

Commented by math khazana by abdo last updated on 04/Aug/18

$$$$$$errorfromline12wehave$$$$h^{{}^{\prime }}\left(x\right)=\frac{\pi }{2x}-\frac{1}{x}\frac{2}{\sqrt{1-x^2}}arctan\sqrt{\frac{1-x}{1+x}}\Rightarrow$$$$h\left(x\right)=\frac{\pi }{2}ln\mid x\mid -\int \frac{2arctan\sqrt{\frac{1-x}{1+x}}}{x\sqrt{1-x^2}}dx+cbut$$$$changementx=costgive$$$$\int \frac{2arctan\sqrt{\frac{1-x}{1+x}}}{x\sqrt{1-x^2}}dx=-\int \frac{2arctan\left(tan\left(\frac{t}{2}\right)\right)}{costsint}sintdt$$$$=-\int \frac{t}{cost}dt{=}_{tan\left(\frac{t}{2}\right)=u}-\int \frac{2arctanu}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$=-\int \frac{arctanu}{1-u^2}duanywaywehave$$$$h\left(x\right)=\frac{\pi }{2}ln\mid x\mid +{\int }_1^x\frac{t}{cost}dt+c$$$$c=h\left(1\right)={\int }_0^{\frac{\pi }{2}}ln(1+cos\theta )d\theta \Rightarrow$$$$h\left(x\right)=\frac{\pi }{2}ln\mid x\mid +{\int }_1^x\frac{t}{cost}dt+{\int }_0^{\frac{\pi }{2}}ln(1+cos\theta )d\theta$$$$if\mid x\mid <1$$$$if\mid x\mid >1wefollowthesameway$$

Commented by math khazana by abdo last updated on 04/Aug/18

$$letfindk\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1-xcos\theta )d\theta$$$$wehaveg\left(x\right)=h\left(x\right)+k\left(x\right)\Rightarrow k\left(x\right)=g\left(x\right)-h\left(x\right)\Rightarrow$$$$k(-x)=g(-x)-h(-x)=g\left(x\right)-h\left(x\right)=k\left(x\right)$$$$becausegandhareevenso$$$$k\left(x\right)=k(-x)=h\left(x\right)\Rightarrow$$$$g\left(x\right)=2h\left(x\right)$$$$$$

Commented by math khazana by abdo last updated on 04/Aug/18

$$g\left(x\right)=\pi ln\mid x\mid +2{\int }_1^x\frac{t}{cost}dt+2{\int }_0^{\frac{\pi }{2}}ln(1+cos\theta )d\theta .$$

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