Definite integration $\int_0^{\pi/2} x^2\csc^2(x)dx$

Question

I am unable to find the definite integral for the given function.

$$\int_0^{\pi/2} x^2\csc^2(x)dx$$

I tried integrating by parts and approaching it like an indefinite integral. But I am convinced that the problem involves some methods of definite integrals.

Answer 1

Although the indefinite integral of $x^2\csc^2x$ is not elementary, its definite integration still is. First, integrate-by-parts

$$\int_0^{\pi/2} x^2\csc^2x\>dx=-\int_0^{\pi/2} x^2d(\cot x)dx =2\int_0^{\pi/2}x\cot x \>dx$$ $$=2\int_0^{\pi/2}xd(\ln \sin x)=-2\int_0^{\pi/2}\ln \sin x\> dx=-2J\tag 1$$

where

$$J=\int_0^{\pi/4}\ln \sin x dx+\int_{\pi/4}^{\pi/2}\ln \sin x dx =\int_0^{\pi/4}(\ln \sin x + \ln\cos x)dx$$ $$=\int_0^{\pi/4}\ln(\frac12\sin2 x )dx =-\frac\pi4 \ln2 + \frac12\int_0^{\pi/2}\ln\sin t dt = -\frac\pi4 \ln2 + \frac12J$$

Thus, $J=-\frac\pi2\ln2$. Substitute $J$ into (1) to get

$$\int_0^{\pi/2} x^2\csc^2x\>dx=\pi\ln2$$

Answer 2

Since you tagged "taylor-expansion", let us try to see what we can do with it.

Start with $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}-\frac{x^8}{315}+O\left(x^{10}\right)$$ $$x^2 \csc ^2(x)=\frac{x^2}{x^2-\frac{x^4}{3}+\frac{2 x^6}{45}-\frac{x^8}{315}+O\left(x^{10}\right)}$$ Now, long division $$x^2 \csc ^2(x)=1+\frac{x ^2}{3}+\frac{x^4}{15}+\frac{2 x^6}{189}+O\left(x^8\right)$$ Integrating termwise $$\int x^2 \csc ^2(x)\,dx=x+\frac{x^3}{9}+\frac{x^5}{75}+\frac{2 x^7}{1323}+O\left(x^9\right)$$ Using the bounds, this would give $$\int_0^{\frac\pi 2}x^2 \csc ^2(x)\,dx\sim \frac{\pi }{2}+\frac{\pi ^3}{72}+\frac{\pi ^5}{2400}+\frac{\pi ^7}{84672} \approx 2.16462$$ while the exact solution is $\pi \log (2)\approx 2.17759$.

Let us see what happens if we add more terms in the base expansion of the sine (say to $O\left(x^{2n+1}\right)$. The numerical values will be $$\left( \begin{array}{cc} n & \text{approximation} \\ 1 & 2.00144 \\ 2 & 2.12895 \\ 3 & 2.16462 \\ 4 & 2.17420 \\ 5 & 2.17671 \\ 6 & 2.17736 \\ 7 & 2.17753 \\ 8 & 2.17757 \\ 9 & 2.17758 \\ 10 & 2.17759 \end{array} \right)$$

Edit

If you use the approximation

$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed more than $1400$ years ago by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician, you will have $$ x^2 \csc ^2(x) \sim \frac{\left(5 \pi ^2-4 (\pi -x) x\right)^2}{256 (\pi -x)^2}$$ Using partial fraction decomposition, then $$ x^2 \csc ^2(x) \sim \frac{x^2}{16}+\frac{5 \pi ^3}{32 (x-\pi )}+\frac{25 \pi ^4}{256 (x-\pi)^2}+\frac{5 \pi ^2}{32}$$ $$\int x^2 \csc ^2(x)\,dx \sim \frac{x^3}{48}+\frac{5 \pi ^2 x}{32}+\frac{25 \pi ^4}{256 (\pi -x)}+\frac{5}{32} \pi ^3 \log (\pi -x)-\frac{17 \pi ^3}{96}$$ Using the bounds $$\int_0^{\frac\pi 2}x^2 \csc ^2(x)\,dx\sim \frac{137-120 \log (2)}{768} \pi ^3 \approx 2.17296$$