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Here is another way: the identity

$$(a+b)^2 - (a-b)^2 = 4ab$$

converts the proposed integral into

$$\frac{1}{4}\left(\int_{-1}^1\,\ln^2(1-x^2)\,dx - \int_{-1}^1\,\ln^2\left(\frac{1+x}{1-x}\right)\,dx\right).$$

Here both integrals are easy to work out.

Thanks, Riemann. The Upper limit should be $\pi/2$.

Evaluate
$$\int_0^{\pi/2}\,\frac{x}{\sin x}\,\log(1 - \sin x)\,dx.$$

For $k \in \mathbb{N}$, let
$$S(k) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+k}\,H_n,$$
where $H_n$ is the $n$-th harmonic number. It is known that
$$S(0) = \frac{\pi^2}{12} - \frac{1}{2}\,\ln^22,\,\,\,\, S(1) = \frac{1}{2}\,\ln^22.$$
Can you find a closed form for $S(k)$ in general?

Show that
$$\sum_{n=1}^\infty\,\frac{(-1)^{n+1}}{(n+1)^2}H_nH_{n+1} = \frac{\pi^4}{480},$$
where $H_n$ is the $n$-th Harmonic number.

For $x > 0$, prove that
$$\left(\sum_{n=0}^\infty\frac{1}{(n+x)^2}\right)^2 \geq 2\,\sum_{n=0}^\infty\frac{1}{(n+x)^3}.$$

Let the proposed series be $f(x)$. Then
$$f'(x) = \sum_{n=1}^\infty (-1)^{n-1}H_{2n}x^{2n} = - \sum_{n=1}^\infty H_{2n}(-x^2)^n.$$
Recall that
$$\sum_{n=1}^\infty H_nz^n = - \frac{\ln(1-z)}{1-z}.$$
It follows that
\begin{eqnarray*}
\sum_{n=1}^\infty H_{2n}z^{2n} & = & \frac{1}{2}\,(\sum_{n=1}^\infty ...

Let the integral be $I_n$. We show that
$$I_n = \left\{\begin{array}{ll}
\frac{1}{2k}
\left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1\right), & \mbox{if $n = 2k+1, k \geq 1$}\\
\frac{2\ln 2}{2k-1} + \frac{1}{2k-1}\left(\frac{1}{2k-2} - \frac{1}{2k-3} + \cdots - 1\right), & \mbox{if $n = 2k, k ...

Recall that
$$\sum_{n=1}^\infty \frac{1}{n}\,x^n = - \ln(1-x).$$
Let $i = \sqrt{-1}$. Then
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n 3^n}\,\cos\left(\frac{n\pi}{12}\right) & = & \Re\left\{\sum_{n=1}^\infty \frac{1}{n}\,\left(\frac{e^{i\pi/12}}{3}\right)^n\right\}\\
& = & \Re\left\{-\ln\left(1 ...

Find a function $f(x)$ which satisfies the functional equation:
$$f^2(x) = c(x^2 + 1) + f(x^2),$$
where $c$ is a positive constant.