# About an integral of the MIT Integration Bee Finals (2023)

I would like to solve the first problem of the MIT Integration Bee Finals, which is the following integral :

$$\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx$$

I tried substitution $$u=\tan(x)$$, King Property, but nothing leads me to the solution which is apparently $$\frac{2\sqrt{3}}{9} \pi$$.

If anybody knows how to solve it I would be grateful.

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$$\int_0^{\frac{\pi}2} \frac{\sqrt[3]{\tan x}}{(1+ \tan x)^2}\frac{dx}{\cos^2x}$$ $$=\int_0^\infty\frac{t^{1/3}}{(1+t)^2}dt=\int_0^\infty\frac s{(1+s^3)^2}3s^2ds,$$ and you certainly know how to integrate a rational fraction.

• I was solving the integral via an alternative, only to arrive at $\displaystyle{\int_0^\infty\frac{t^{1/3}}{(1+t)^2}dt}$, hence I posted it as brief comment underneath this answer: $$\int_0^{\frac{\pi}{2}} \dfrac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx = \int_0^{\frac{\pi}{2}} \dfrac{\sqrt[3]{\tan(x)}}{1 + \sin{2x}}dx \quad \xrightarrow{\large{t \space = \space \tan{x}}} \quad \int_0^\infty \dfrac{t^{1/3}}{1 + \dfrac{2t}{1 + t^2}} \cdot \dfrac{1}{1 + t^2} dt = \int_0^\infty \dfrac{t^{1/3}}{(1+t)^2}dt$$ Feb 26, 2023 at 11:19

$$u = \tan^\frac{1}{3}x,\\ 3~\mathrm du = \tan^\frac{-2}{3}x \sec^2x ~\mathrm dx$$

Shortens it down to -

$$\int \frac{3u^3~\mathrm du}{(1+u^3)^2}$$

Split it into $$\displaystyle\int \frac{u \cdot 3u^2~\mathrm du}{(1+u^3)^2}$$ and perform by parts.

• (+1) I appreciate the last line! Feb 25, 2023 at 18:18

I want to generalize the integral as $$I(a,n)=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(\sin x+\cos x)^{2 n}}=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(1+\sin 2 x)^n} d x$$ where $$0 and $$n\in N$$.

Letting $$t=\tan x$$ gives $$I(a,n)=\int_0^{\infty} \frac{t^{a}}{\left(1+\frac{2 t}{1+t^2}\right)^n} \cdot \frac{d t}{1+t^2}=\int_0^{\infty} \frac{t^{a}\left(1+t^2\right)^{n-1}}{(t+1)^{2 n}} d t$$ Using Binomial expansion and beta function, we have \begin{aligned} I(a,n)&=\sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \int_0^{\infty} \frac{t^{2 k+a}}{(t+1)^{2 n}} d t\\&=\sum_{k=0}^{n-1}\left(\begin{array}{l} n \\ k \end{array}\right) B\left(2 k+a+1,2n-2k-a-1 \right)\\&= \sum_{k=0}^{n-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) \frac{\Gamma\left(2 k+a+1\right) \Gamma\left(2 n-2 k—a-1\right)}{(2 n-1)!}\end{aligned}

In particular, $$I(\frac{1}{3} ,1)=\Gamma(\frac{4}{3} )\Gamma(\frac{2}{3} )= \frac{1}{3} \Gamma(\frac{1}{3} )\Gamma(\frac{2}{3} ) =\frac{2\pi}{3\sqrt 3}$$

\begin{align*} \textbf{I}&=\int \limits_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{(\sin x+\cos x)^{2}}\,\mathrm{d}x =\int \limits_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{\{\cos x(\tan x+1)\}^2}\,\mathrm{d}x=\int \limits_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{\cos^2 x(\tan x+1)^2}\,\mathrm{d}x\\ &=\int \limits_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{ (\tan x+1)^2}\,\sec^{2}x\mathrm{d}x \end{align*} Substitute $$\tan x=u\Rightarrow \sec^{2}x\mathrm{d}x=\mathrm{d}u$$. When $$x\rightarrow \frac{\pi}{2}, u\rightarrow \infty$$ and $$x\rightarrow 0, u\rightarrow 0$$. It follows, $$\textbf{I}=\int\limits_{0}^{\infty} \frac{\sqrt[3]{u}}{(u+1)^2}\,\mathrm{d}u$$ Substitute $$u=z^{3}\Rightarrow \mathrm{d}u=3z^{2}\mathrm{d}z$$. When $$u\rightarrow \infty, z\rightarrow \infty$$ and $$u\rightarrow 0, z\rightarrow 0$$. Follow on, $$\textbf{I}=\int\limits_0^\infty \frac{z}{(z^3+1)^2}3z^2\,\mathrm{d}z=\int\limits_0^\infty \frac{3z^3}{(z^3+1)^2}\,\mathrm{d}z=\int\limits_0^\infty \frac{3z^3}{(z+1)^2(z^2-z+1)^2}\,\mathrm{d}z$$ By employing partial fractional decomposition and using the resulting expressions, we can solve the integral and continue with subsequent calculations. $$\frac{3z^3}{(z+1)^2(z^2-z+1)^2}\equiv \frac{A}{z+1}+\frac{B}{(z+1)^2}+\frac{Cz+D}{z^2-z+1}+\frac{Ez+F}{(z^2-z+1)^2}$$ Multiply both sides of the equivalency by $$(z+1)^2(z^2-z+1)^2$$, $$\begin{equation*} \begin{split} 3z^3 \equiv A(z^3+1)(z^2-z+1) &+ B(z^2-z+1)^2 \\ &+ (Cz+D)(z+1)(z^3+1) + (Ez+F)(z+1)^2 \end{split} \end{equation*}$$ Substitute $$z=-1$$, it follows $$B=-\frac{1}{3}$$. Rewrite: $$\begin{equation*} \begin{split} 3z^3 \equiv (A+C)z^5&+\left(D+C-\frac{1}{3}-A\right)z^4+\left(A+\frac{2}{3}+D+E\right)z^3\\ &+(A-1+C+2E+F)z^2 +\left(2F+E+D+C+\frac{2}{3}-A\right)z\\ &+\left(A-\frac{1}{3}+D+F\right) \end{split} \end{equation*}$$ By equating the coefficients of $$z^5,z^4,z^3,z^2,z,1$$ the following system of equations in found: $$\begin{equation*} \begin{cases} A+C=0\\ D+C-\dfrac{1}{3}-A=0\\ A+\dfrac{2}{3}+D+E=3\\ A-1+C+2E+F=0\\ 2F+E+D+C+\dfrac{2}{3}-A=0\\ A-\dfrac{1}{3}+D+F=0 \end{cases} \end{equation*}$$ Solving the equation the values are found. $$A=\dfrac{1}{3},B=-\dfrac{1}{3},C=-\dfrac{1}{3},D=1,E=1,F=-1$$. Substituting the value into the equivalency, it follows:
$$\frac{3z^3}{(z^3+1)^2}=\frac{1}{3(z+1)}-\frac{1}{3(z+1)^2}-\boxed{\frac{z-3}{3(z^2-z+1)}}+\boxed{\frac{z-1}{(z^2-z+1)^2}}$$ Expand the boxed fractions: $$\begin{equation*} \begin{split} \frac{3z^3}{(z^3+1)^2}=\frac{1}{3(z+1)}&-\frac{1}{3(z+1)^2}-\boxed{\frac{2z-1}{6(z^2-z+1)}+\frac{5}{6(z^2-z+1)}}\\ &+\boxed{\frac{2z-1}{2(z^2-z+1)^2}-\frac{1}{2(z^2-z+1)^2}} \end{split} \end{equation*}$$ Rewrite $$(z^2-z+1)$$ as $$(z-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2$$: $$\begin{equation*} \begin{split} \frac{3z^3}{(z^3+1)^2}=\frac{1}{3(z+1)}&-\frac{1}{3(z+1)^2}-\boxed{\frac{2z-1}{6(z^2-z+1)}+\frac{5}{6\left\{(z-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2\right\}}}\\ &+\boxed{\frac{2z-1}{2(z^2-z+1)^2}-\frac{1}{2\left\{(z-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2\right\}^2}} \end{split} \end{equation*}$$ Integrate both sides of the identity: $$\begin{equation*} \begin{split} \int\frac{3z^3}{(z^3+1)^2}\,\mathrm{d}x=\int \frac{1}{3(z+1)}&-\int \frac{1}{3(z+1)^2}-\int {\frac{2z-1}{6(z^2-z+1)}+\int \frac{5}{6\left\{(z-\frac{1}{2})^2+ (\frac{\sqrt{3}}{2})^2\right\}}}\\ &+\int {\frac{2z-1}{2(z^2-z+1)^2}-\int \frac{1}{2\left\{(z-\frac{1}{2})^2+ (\frac{\sqrt{3}}{2})^2\right\}^2}}\,\mathrm{d}x \end{split} \end{equation*}$$ $$=\frac{1}{3}\ln(z+1) +\frac{1}{3(z+1)} -\frac{1}{6}\ln(z^2-z+1)+\frac{5}{3\sqrt{3}}\tan^{-1}\left(\frac{2z-1}{\sqrt{3}}\right)-\frac{1}{2(z^2-z+1)}$$ $$-\frac{2}{3\sqrt{3}}\tan^{-1}\left(\frac{2z-1}{\sqrt{3}}\right)-\frac{2z-1}{6(z^2-z+1)}$$ Therefore, \begin{align*} \int\frac{3z^3}{(z^3+1)^2}\,\mathrm{d}x&=\frac{1}{6}\ln \left(\frac{z^2+2z+1}{z^2-z+1}\right)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z-1}{\sqrt{3}}\right)-\frac{z}{z^3+1}+C\\ &=\frac{1}{6}\ln \left(\frac{1+\frac{2}{z}+\frac{1}{z^2}}{1-\frac{1}{z}+\frac{1}{z^2}}\right)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z-1}{\sqrt{3}}\right)-\frac{\frac{1}{z^2}}{1+\frac{1}{z^3}}+C \end{align*} Using the Fundamental theorem of calculus, it follows: \begin{align*} \int\limits_0^\infty \frac{3z^3}{(z^3+1)^2}\,\mathrm{d}x&=\left[\frac{1}{6}\ln \left(\frac{1+\frac{2}{z}+\frac{1}{z^2}}{1-\frac{1}{z}+\frac{1}{z^2}}\right)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z-1}{\sqrt{3}}\right)-\frac{\frac{1}{z^2}}{1+\frac{1}{z^3}}\right]_{0}^{\infty}\\ &=\frac{1}{\sqrt{3}}\frac{\pi}{2}+\frac{1}{\sqrt{3}}\frac{\pi}{6}=\frac{2\pi}{3\sqrt{3}} \end{align*} Therefore $$\int \limits_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{(\sin x+\cos x)^{2}}\,\mathrm{d}x=\frac{2\pi}{3\sqrt{3}}$$

There are branch-cut issues but I tried this in the given short time (assuming I am in the exam):

Let $$I=\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx=\int_0^\infty\frac{z^{1/3}}{(1+z)^2}dz=\int_0^\infty\frac{2z^{5/3}}{(1+z^2)^2}dz$$. Then by using the contour integration along the famous counter-clockwise closed semi-circle contour on the upper half plane and the residue theorem: $$\int_{-\infty}^\infty\frac{2z^{5/3}}{(1+z^2)^2}dz=(e^{5\pi i/3}+1)I=2\pi iRes_{z=i}\frac{2z^{5/3}}{(1+z^2)^2}=2\pi i(-\frac13e^{\pi i/3})$$ Hence, $$(e^{2\pi i/3}+e^{\pi i/3})I=2\pi i/3\implies I=\frac{2\pi}{3\sqrt 3}.$$

Note: Residue calculation

• @AnneBauval Thanks Feb 26, 2023 at 11:50