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Trigonometric logarithmic integral

Here is another integral that involves both trigonometric and logarithmic functions as the previous one. The problem is stated as follows:

Evaluate the integral

    \[\mathcal{J} = \int_0^{\pi/2 }\sin^2x \log  \left(\sin^2(\tan x)\right) \, {\rm d}x\]

Solution

Well, the obvious move to make at this particular problem is to make the sub. u=\tan x. Hence:

    \begin{align*} \mathcal{J} &= \int_{0}^{\pi/2} \sin^2 x \log \left ( \tan^2 x \right ) \, {\rm d}x \\ &\!\!\!\!\!\!\overset{u=\tan x}{=\! =\! =\! =\!} 2\int_{0}^{\infty} \frac{u^2}{\left ( 1+u^2 \right )^2} \log \left ( \sin u \right ) \, {\rm d}u\\ &=2 \int_{0}^{\infty} \frac{u^2}{\left ( 1+u^2 \right )^2} \left ( - \log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nu}{n} \right ) \, {\rm d}u \\ &= -2 \log 2 \int_{0}^{\infty} \frac{u^2}{\left ( 1+u^2 \right )^2} \, {\rm d}u - \\ & \quad \quad -2 \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{u^2 \cos 2nu}{\left ( 1+u^2 \right )^2} \, {\rm d }u\\ &= - \frac{\pi \log 2}{2} -\frac{1}{2} \sum_{n=1}^{\infty} e^{-2n} \left ( \pi - 2 \pi n \right ) \\ &= - \frac{\pi \log 2}{2} + \frac{\pi}{e^2-1} - \frac{\pi}{2} \left [ 2 - \log \left ( e^2-1 \right ) \right ] \\ \end{align*}

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