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Question Number 91303 by 174 last updated on 29/Apr/20

Commented by mathmax by abdo last updated on 29/Apr/20

$$A={\int }_0^{\frac{\pi }{2}}ln(1+{cos}^2\left(2x\right))dx\Rightarrow A={\int }_0^{\frac{\pi }{2}}ln(1+\frac{1+cos\left(4x\right)}{2})dx$$$$={\int }_0^{\frac{\pi }{2}}ln(3+cos\left(4x\right))dx-\frac{\pi }{2}ln\left(2\right)wehave$$$${\int }_0^{\frac{\pi }{2}}ln(3+cos\left(4x\right))dx=\frac{\pi }{2}ln\left(3\right)+{\int }_0^{\frac{\pi }{2}}ln(1+\frac{1}{3}cos\left(4x\right))dx(4x=t)$$$$=\frac{\pi }{2}ln\left(3\right)+\frac{1}{4}{\int }_0^{\pi }ln(1+\frac{1}{3}cost)\Rightarrow$$$$A=\frac{\pi }{2}ln\left(3\right)-\frac{\pi }{2}ln\left(2\right)+\frac{1}{4}{\int }_0^{\pi }ln(1+\frac{1}{3}cost)dtlet$$$$f\left(a\right)={\int }_0^{\pi }ln(1+acost)dtwith0<a<1$$$$wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^{\pi }\frac{cost}{1+acost}dt=\frac{1}{a}{\int }_0^{\pi }\frac{1+acost-1}{1+acost}dt$$$$=\frac{\pi }{a}-\frac{1}{a}{\int }_0^{\pi }\frac{dt}{1+acost}wehave$$$${\int }_0^{\pi }\frac{dt}{1+acost}{=}_{tan\left(\frac{t}{2}\right)=u}{\int }_0^{\mathrm{\infty }}\frac{2du}{(1+u^2)(1+a\times \frac{1-u^2}{1+u^2})}$$$$={\int }_0^{\mathrm{\infty }}\frac{2du}{1+u^2+a-{au}^2}={\int }_0^{\mathrm{\infty }}\frac{2du}{(1-a)u^2+1+a}=\frac{2}{1-a}{\int }_0^{\mathrm{\infty }}\frac{du}{u^2+\frac{1+a}{1-a}}$$$${=}_{u=\sqrt{\frac{1+a}{1-a}}z}\frac{2}{1-a}{\int }_0^{\mathrm{\infty }}\frac{1}{\frac{1+a}{1-a}(1+z^2)}\times \sqrt{\frac{1+a}{1-a}}dz$$$$=\frac{2}{\sqrt{1-a^2}}{\int }_0^{\mathrm{\infty }}\frac{dz}{1+z^2}=\frac{2}{\sqrt{1-a^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-a^2}}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\frac{\pi }{a}-\frac{\pi }{a\sqrt{1-a^2}}\Rightarrow f\left(a\right)=\pi ln\left(a\right)-\pi \int \frac{da}{a\sqrt{1-a^2}}+c$$$$\int \frac{da}{a\sqrt{1-a^2}}{=}_{a=sinx}\int \frac{cosxdx}{sinx\times cosx}=\int \frac{dx}{sinx}$$$${=}_{tan\left(\frac{x}{2}\right)=u}\int \frac{2du}{(1+u^2)\times \frac{2u}{1+u^2}}=\int \frac{du}{u}=ln\mid u\mid =ln\mid tan\left(\frac{x}{2}\right)\mid$$$$=ln\mid tan\left(\frac{1}{2}arcsina\right)\mid \Rightarrow$$$$f\left(a\right)=\pi ln\left(a\right)-\pi ln\mid tan\left(\frac{1}{2}arcsina\right)\mid +c$$$$f\left(a\right)=\pi ln\left(\frac{a}{\mid tan\left(\frac{1}{2}arcsina\right)\mid }\right)+c$$$$\frac{1}{2}arcsina\sim \frac{a}{2}\Rightarrow tan\left(\frac{arcsina}{2}\right)\sim \frac{a}{2}\Rightarrow f\left(a\right)\sim \pi ln\left(2\right)+c(a\sim 0)$$$$\Rightarrow {lim}_{a\to 0}f\left(a\right)=0=\pi ln\left(2\right)+c\Rightarrow c=-\pi ln\left(2\right)\Rightarrow$$$$f\left(a\right)=\pi ln\left(a\right)-\pi ln\left(2\right)-\pi ln\mid tan\left(\frac{1}{2}arcsina\right)\mid so$$$${\int }_0^{\pi }ln(1+\frac{1}{3}cost)dt=f\left(\frac{1}{3}\right)=-\pi ln\left(3\right)-\pi ln\left(2\right)$$$$-\pi ln\mid tan(\frac{1}{2}arcsin\left(\frac{1}{3}\right)\mid \Rightarrow$$$$A=\frac{\pi }{2}ln\left(3\right)-\frac{\pi }{2}ln\left(2\right)-\frac{\pi }{4}ln\left(3\right)-\frac{\pi }{4}ln\left(2\right)-\frac{\pi }{4}ln\mid tan(\frac{1}{2}arcsin\left(\frac{1}{3}\right)\mid$$$$A=\frac{\pi }{4}ln\left(3\right)-\frac{3\pi }{4}ln\left(2\right)-\frac{\pi }{4}ln\mid tan(\frac{1}{2}arcsin\left(\frac{1}{3}\right)\mid$$

Commented by 174 last updated on 30/Apr/20

thanks alot

Commented by mathmax by abdo last updated on 30/Apr/20

$$youarewelcome$$

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