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Question Number 91493 by Zainal Arifin last updated on 01/May/20

$$\underset{0}{\stackrel{\pi /2}{\int }}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx=$$

Commented by mathmax by abdo last updated on 01/May/20

$$I={\int }_0^{\frac{\pi }{2}}ln\left(\frac{1+\frac{3}{4}sinx}{1+\frac{3}{4}cosx}\right)={\int }_0^{\frac{\pi }{2}}ln(1+\frac{3}{4}sinx)-{\int }_0^{\frac{\pi }{2}}ln(1+\frac{3}{4}cosx)$$$$letf\left(a\right)={\int }_0^{\frac{\pi }{2}}ln(1+asinx)with0<a<1$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+asinx}dx=\frac{1}{a}{\int }_0^{\frac{\pi }{2}}\frac{1+asinx-1}{1+asinx}dx$$$$=\frac{\pi }{2a}-\frac{1}{a}{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+asinx}but{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+asinx}{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^1\frac{2dt}{(1+t^2)(1+a\frac{2t}{1+t^2})}$$$$=2{\int }_0^1\frac{dt}{1+t^2+2at}=2{\int }_0^1\frac{dt}{t^2+2at+1}=2{\int }_0^1\frac{dt}{{(t+a)}^2+1-a^2}$$$${=}_{t+a=\sqrt{1-a^2}u}2{\int }_{\frac{a}{\sqrt{1-a^2}}}^{\frac{1+a}{\sqrt{1-a^2}}}\frac{\sqrt{1-a^2}du}{(1-a^2)(1+u^2)}$$$$=\frac{2}{\sqrt{1-a^2}}(arctan\left(\frac{1+a}{\sqrt{1-a^2}}\right)-arctan\left(\frac{a}{\sqrt{1-a^2}}\right))\Rightarrow$$$$f\left(a\right)=\frac{\pi }{2}ln\left(a\right)-2\int \frac{1}{a\sqrt{1-a^2}}arctan\left(\frac{1+a}{\sqrt{1-a^2}}\right)da$$$$+2\int \frac{1}{a\sqrt{1-a^2}}arctan\left(\frac{a}{\sqrt{1-a^2}}\right)+cwehave$$$$\int \frac{1}{a\sqrt{1-a^2}}arctan\left(\frac{a}{\sqrt{1-a^2}}\right){=}_{a=sinx}\int \frac{1}{sinxcosx}arctan\left(\frac{sinx}{cosx}\right)cosxdx$$$$=\int \frac{x}{sinx}dx$$$$\int \frac{1}{a\sqrt{1-a^2}}arctan\left(\frac{1+a}{\sqrt{1-a^2}}\right)da{=}_{a=cosx}-\int \frac{1}{cosxsinx}arctan\left(\frac{1+cosx}{sinx}\right)sinxdx$$$$=-\int \frac{1}{cosx}arctan\left(\frac{2{cos}^2\left(\frac{x}{2}\right)}{2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}\right)dx$$$$=-\int \frac{1}{cosx}arctan\left(\frac{1}{tanx}\right)dx=-\int \frac{1}{cosx}(\frac{\pi }{2}-x)$$$$=-\frac{\pi }{2}\int \frac{dx}{cosx}+\int \frac{x}{cosx}dx....becontinued...$$$$$$

Answered by MJS last updated on 01/May/20

$$\ln \frac{4+3\sin x}{4+3\cos x}=-\ln \frac{4+3\sin (\frac{\pi }{2}-x)}{4+3\cos (\frac{\pi }{2}-x)}\Rightarrow$$$$\Rightarrow \mathrm{answer}\mathrm{is}0$$

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