Inverse Sine integral

First I will rewrite the integral as $\int_0^{\frac{\pi}{2}} x^2 \csc^2 x dx$ . The form looks naturally vulnerable to integration by parts so the obvious course of action is letting $u = x^2, du=2x, dv = \csc^2 x, v = -\cot x$ to get $-x^2 \cot x \big|_{0}^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 2x \cot x dx$ .

Before tackling the new integral, I will prove that the expression outside of the integral evaluates to 0. We know that $\cot \left( \frac{\pi}{2} \right) = 0$ so that part goes to 0 so we only need to check $\lim_{x \to 0} -x^2 \cot x = \lim_{x \to 0} \frac{-x^2}{\tan x} = \lim_{x \to 0} \frac{-2x}{\sec^2 x} = \frac{2 \times 0}{1} = 0$ by applying L'hopital's rule.

For the new integral I will do integration by parts again, applying the property that $\int \cot x dx = \log \sin x$ let $u = 2x, du = 2, dv = \cot x, v = \log \sin x$ . Then the integral is equal to $2x \log \sin x \big|_{0}^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} 2 \log \sin x dx$ .

Again, I will prove that the bit outside of the integral evaluates to 0. First notice that $\log \sin \frac{\pi}{2} = \log 1 = 0$ . And $\lim_{x \to 0} 2x \log \sin x = \lim_{x \to 0} 2 \frac{\log \sin x}{x^{-1}} = \lim_{x \to 0} 2\frac{ \cot x}{-x^{-2}} dx = \lim_{x \to 0} -2x^2 \cot x = \lim_{x \to 0} -2x^2( x^{-1} + O(x) ) = \lim_{x \to 0} -2x + O(x^3) = 0$ by applying L'hopital's rule first and then using the Laurent series for the cotangent function.

So after all that the integral is actually equal to $-2 \int_0^{\frac{\pi}{2}} \log \sin x dx$ . Now, the integral for $\log \sin x$ over this interval is a familiar one so I will assume that the reader knows, can deduce, or look up on the internet that it evaluates to $-\frac{1}{2} \pi \log 2$ so by multiplying the outside factor our integral evaluates to precisely $\pi \log 2$ .