Strange mistake when calculate improper integral

Question

I want to calculate the following improper integral, so I do: $$ \int_0^{{\pi}/{2}} \cos^{2/3} x \sin^{-2/3} x\,dx = \int_0^{{\pi}/{2}} \frac{\cos x \cos^ {2/3} x}{\cos x \sin^{2/3}x}\,dx = \int_0^{{\pi}/{2}}\frac{3\cos^{2/3} x\,d(\sin^{1/3} x)}{\cos x} =$$ $$\int_0^1\frac{3\,dt}{\sqrt{1-t^2}} = 3\arcsin t |_0^1 = \frac{3\pi}{2} $$

However, when I evalute it numerically in MatLab, it shows the answer about $\pi$. It seems that there is a problem at discontinuity point $x = 0$, but I don't understand why, because I could use limits and get the same result.

Answer 1

The following method of calculation is conceivably the most elementary, though it has the disadvantage of being a good bit longer than more advanced methods (e.g., via the beta function).

Substituting first $t=\cot{(x)}$ followed by $u=t^{\frac13}$, and then expanding by partial fractions, yields:

$$\begin{align} \int_{0}^{\frac{\pi}{2}}\cot^{\frac23}{(x)}\,\mathrm{d}x &=\int_{0}^{\infty}\frac{t^{\frac23}}{t^2+1}\,\mathrm{d}t\\ &=\int_{0}^{\infty}\frac{3u^4}{u^6+1}\,\mathrm{d}u\\ &=\int_{0}^{\infty}\left(\frac{1}{u^2+1}+\frac{2u^2-1}{u^4-u^2+1}\right)\,\mathrm{d}u\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}+\int_{0}^{\infty}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u.\\ \end{align}$$

The second integral in the last line above may be reduced from an integral over $[0,\infty)$ to an integral over $[0,1]$ as follows:

$$\begin{align} \int_{0}^{\infty}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u &=\int_{0}^{1}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u+\int_{1}^{\infty}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u\\ &=\int_{0}^{1}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u+\int_{0}^{1}\frac{2-w^2}{w^4-w^2+1}\,\mathrm{d}w;~\left(\text{from }u\rightarrow\frac{1}{w}\right)\\ &=\int_{0}^{1}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u+\int_{0}^{1}\frac{2-u^2}{u^4-u^2+1}\,\mathrm{d}u\\ &=\int_{0}^{1}\frac{u^2+1}{u^4-u^2+1}\,\mathrm{d}u.\\ \end{align}$$

We can show via substitution that $\int_{0}^{1}\frac{u^2+1}{u^4-u^2+1}\,\mathrm{d}u=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}$. To demonstrate this, we can start from the integral $\int_{0}^{\infty}\frac{\mathrm{d}x}{x^2+1}$ and substitute $\frac{u}{1-u^2}$:

$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{x^2+1} &=\int_{0}^{1}\frac{1}{\left(\frac{u}{1-u^2}\right)^2+1}\cdot\frac{u^2+1}{\left(1-u^2\right)^2}\,\mathrm{d}u\\ &=\int_{0}^{1}\frac{u^2+1}{u^2+\left(1-u^2\right)^2}\,\mathrm{d}u\\ &=\int_{0}^{1}\frac{u^2+1}{u^4-u^2+1}\,\mathrm{d}u.\\ \end{align}$$

Putting this all together, we get:

$$\begin{align} \int_{0}^{\frac{\pi}{2}}\cot^{\frac23}{(x)}\,\mathrm{d}x &=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}+\int_{0}^{\infty}\frac{2u^2-1}{u^4-u^2+1}\,\mathrm{d}u\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}+\int_{0}^{1}\frac{u^2+1}{u^4-u^2+1}\,\mathrm{d}u\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}+\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}\\ &=2\int_{0}^{\infty}\frac{\mathrm{d}u}{u^2+1}\\ &=2\cdot\frac{\pi}{2}=\pi. \end{align}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large% \int_{0}^{\pi/2}\cos^{2/3}\pars{x}\sin^{-2/3}\pars{x}\,\dd x}} ^{\dsc{\sin\pars{x}=t\ \imp\ x=\arcsin\pars{x}}}\ =\ \int_{0}^{1}\pars{1 - t^{2}}^{1/3}t^{-2/3}\,{\dd t \over \pars{1 - t^{2}}^{1/2}} \\[5mm]&=\ \overbrace{\int_{0}^{1}t^{-2/3}\pars{1 - t^{2}}^{-1/6}\,\dd t} ^{\dsc{t^{2}\ \mapsto\ t}}\ =\ \int_{0}^{1}t^{-1/3}\pars{1 - t}^{-1/6}\,\half\,t^{-1/2}\,\dd t \\[5mm]&=\half\int_{0}^{1}t^{-5/6}\pars{1 - t}^{-1/6}\,\dd t =\half\,{\Gamma\pars{1/6}\Gamma\pars{5/6} \over \Gamma\pars{1}} =\half\,{\pi \over \sin\pars{\pi/6}}=\half\,{\pi \over 1/2} =\color{#66f}{\Large\pi} \end{align}