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I want to evaluate the value of the following integral: $$\int_0^\frac{\pi}{2}\frac{1}{\sin^2{x}}\ln{\frac{1+a*\sin^2{x}}{1-a*\sin^2{x}}}dx$$

i have tried 2 methods, but fail to proceed.

1. \begin{align*} I&=\int_0^\frac{\pi}{2}\frac{1}{\sin^2{x}}\ln{\frac{1+a*\sin^2{x}}{1-a*\sin^2{x}}}dx\\ \frac{dI}{da}&=\int_0^\frac{\pi}{2}\frac{2}{1-a^2*\sin^4{a}}dx\\ &=\int_0^\frac{\pi}{2}\frac{1}{1-a*\sin^2{x}}+\frac{1}{1+a*\sin^2{x}}dx \end{align*}

2. using $u=\ln{\frac{1+a*\sin^2{x}}{1-a*\sin^2{x}}}$ and $dv=\frac{dx}{\sin^2{x}}$ we have \begin{align*} I&=\bigg[-\cot{x}\ln{\frac{1+a*\sin^2{x}}{1-a*\sin^2{x}}}\bigg]_0^\frac{\pi}{2}+4a\int_0^\frac{\pi}{2}\frac{\cos^2{x}}{1-a^2*\sin^4{x}}dx\\ &=4a\int_0^\frac{\pi}{2}\frac{\cos^2{x}}{1-a^2*\sin^4{x}}dx \end{align*} which is similar to the first one

Reynan Henry
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2 Answers2

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Try this:

$$t=\cot x;\quad dt=-\frac{dx}{\sin^2 x};\frac{1}{\sin^2 x}=1+t^2$$

$$I=\int_0^{\pi/2}\frac{1}{\sin^2 x}\ln\frac{1+a\sin^2 x}{1-a\sin^2 x}dx$$ $$I=-\int_{\infty}^{0}\ln\frac{1+a\sin^2 x}{1-a\sin^2 x}dt= \int_0^\infty\ln\frac{1/\sin^2 x+a}{1/\sin^2 x-a}dt= \int_0^\infty\ln\frac{1+t^2+a}{1+t^2-a}dt $$

Now you can consider separately just the indefinite integral $\int\ln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).

orion
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Considering $$I=\int\frac{dx}{1-a \sin ^2(x)}$$ let $$\tan(x)=t \implies x=\tan ^{-1}(t) \implies dx=\frac{dt}{1+t^2}$$ which make $$I=\int\frac{dt}{1-(a-1) t^2}=\frac{\tanh ^{-1}\left(\sqrt{a-1} t\right)}{\sqrt{a-1}}$$ Just do the same for the other

Claude Leibovici
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