Integral $\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx$

Question

I am trying to compute the integral

$$\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx \tag{1}$$

Context: Originally I was trying to prove the following result:

$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})\tag{2}$$

Where $\beta(2)$ is the Catalan´s constant

To this end I started with the well known result

$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{4^n x^{2n+1}}{(2n+1)\binom{2n}{n}} \tag{3}$$

Dividing both sides of $(3)$ by $x$ and integrating from $0$ to $1/2$ we obtain

$$\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=\frac12\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}} \tag{4}$$

So the task reduces to compute the integral in $(4)$. Therefore

$$ \begin{aligned} \sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}}&=2\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx\\ &=2\int_0^{\pi/6}\frac{x}{\sin(x)}\,dx &(x \to \sin(x))\\ &=2\int_0^{\pi/2}\frac{x}{\sin(x)}\,dx-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\ &=4\beta(2)-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\ &=4\beta(2)-2\int_{0}^{\pi/3}\frac{\left(\frac{\pi}{2}-x\right)}{\cos(x)}\,dx & (x \to \frac{\pi}{2}-x)\\ &=4\beta(2)-\pi\int_{0}^{\pi/3}\sec(x)\,dx+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\ &=4\beta(2)-\pi\ln\left(\sec(x)+\tan(x) \right)\Big|_0^{\pi/3}+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\ &=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\ &=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{x}{e^{ix}+e^{-ix}}\,dx\\ &=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{xe^{-ix}}{1+e^{-2ix}}\,dx\\ &=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}xe^{-ix}\sum_{k=0}^\infty(-1)^ke^{-2ikx}\,dx\\ &=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\sum_{k=0}^\infty(-1)^k\int_{0}^{\pi/3}xe^{-ix(2k+1)}\,dx\\ \end{aligned} $$

The integral in the last line is $(1)$. I integrated by parts, but ended up with some nasty series not very promising.

Answer 1

$$I=\int_0^\frac{\pi}{3}\frac{x}{\cos(x)}\,dx $$ Making the substitution $x=\frac{\pi}{2}-t$ $$I=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{dt}{\sin t}-\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{t\,dt}{\sin t}=I_1-I_2$$ Next, we make the substitution $\frac{dt}{\sin t}=d\big(\ln(\tan\frac{t}{2})\big)$ $$I_1=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2}d\big(\ln(\tan\frac{t}{2})\big)=-\frac{\pi}{2}\ln(\tan\frac{\pi}{12})$$ Using $\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$ and $-\ln(2-\sqrt 3)=\ln(2+\sqrt 3)$ $$I_1=\frac{\pi}{2}\ln(2+\sqrt3)$$ $$I_2=\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{t\,dt}{\sin t}=t\ln(\tan\frac{t}{2})\Big|_\frac{\pi}{6}^\frac{\pi}{2}-\int_\frac{\pi}{6}^\frac{\pi}{2}\ln(\tan\frac{t}{2})dt$$ $$=\frac{\pi}{6}\ln(2+\sqrt 3)-\int_0^\frac{\pi}{2}\ln(\tan\frac{t}{2})dt+\int_0^\frac{\pi}{6}\ln(\tan\frac{t}{2})dt$$ $$=\frac{\pi}{6}\ln(2+\sqrt 3)-2\int_0^\frac{\pi}{4}\ln(\tan x)dx+2\int_0^\frac{\pi}{12}\ln(\tan x)dx$$ In the second term we make the substitution $\tan x=t$, and the third term was evaluated here

($G=\beta(2)$ - Catalan's constant) $$I_2=\frac{\pi}{6}\ln(2+\sqrt 3)-2\int_0^1\frac{\ln t}{1+t^2}dt -\frac{4}{3}G=\frac{\pi}{6}\ln(2+\sqrt 3)+2G-\frac{4}{3}G$$ $$I_2=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac{2}{3}G$$ $$I=I_1-I_2=\frac{\pi}{3}\ln(2+\sqrt 3)-\frac{2}{3}G$$

Answer 2

Let $x=2\arctan(y),$ hence $\mathrm{d}x=\frac{2\,\mathrm{d}y}{y^2+1},$ and $\cos(x)=\cos[\arctan(y)]^2-\sin[\arctan(y)]^2=\frac1{y^2+1}-\frac{y^2}{y^2+1}=-\frac{y^2-1}{y^2+1},$ which means $$\int_0^{\pi/3}\frac{x}{\cos(x)}\,\mathrm{d}x=\int_0^{\frac1{\sqrt{3}}}2\arctan(y)\cdot\left[-\frac{y^2+1}{y^2-1}\right]\frac2{y^2+1}\,\mathrm{d}y$$ $$=-2\int_0^{\frac1{\sqrt{3}}}\arctan(y)\frac2{y^2-1}\,\mathrm{d}y=-2\int_0^{\frac1{\sqrt{3}}}\arctan(y)\frac{(y+1)-(y-1)}{y^2-1}\,\mathrm{d}y$$ $$=-2\int_0^{\frac1{\sqrt{3}}}\arctan(y)\left[\frac1{y-1}-\frac1{y+1}\right]\,\mathrm{d}y$$ $$=2\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y-2\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y-1}\,\mathrm{d}y$$ $$=2\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y+2\int_0^{-\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y.$$ Now, $$\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y=\ln\left(1+\frac1{\sqrt{3}}\right)\arctan\left(-\frac1{\sqrt{3}}\right)-\int_0^{\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y$$ $$\int_0^{-\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y=-\ln\left(1-\frac1{\sqrt{3}}\right)\arctan\left(\frac1{\sqrt{3}}\right)-\int_0^{-\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y$$ hence $$2\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y+2\int_0^{-\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y-2\int_0^{-\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y-2\int_0^{-\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y.$$ Notice that $$-2\int_0^{-\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y=2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(1-y)}{y^2+1}\,\mathrm{d}y,$$ so $$2\int_0^{\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y+2\int_0^{-\frac1{\sqrt{3}}}\frac{\arctan(y)}{y+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y+2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(1-y)}{y^2+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(y+1)}{y^2+1}\,\mathrm{d}y-2\int_0^{\frac1{\sqrt{3}}}\frac{-\ln(1-y)}{y^2+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac{\ln(1+y)-\ln(1-y)}{y^2+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac1{y^2+1}\sum_{m=1}^{\infty}\frac{y^m-(-y)^m}{m}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac1{y^2+1}\sum_{m=0}^{\infty}\frac{y^{2m+1}}{2m+1}\,\mathrm{d}y.$$ Let $z=\arctan(y),$ so $$-2\int_0^{\frac1{\sqrt{3}}}\frac1{y^2+1}\sum_{m=0}^{\infty}\frac{y^{2m+1}}{2m+1}\,\mathrm{d}y=-2\int_0^{\pi/6}\sum_{m=0}^{\infty}\frac{\tan(z)^{2m+1}}{2m+1}\,\mathrm{d}z,$$ so $$\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\frac1{\sqrt{3}}}\frac1{y^2+1}\sum_{m=0}^{\infty}\frac{y^{2m+1}}{2m+1}\,\mathrm{d}y$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\int_0^{\pi/6}\sum_{m=0}^{\infty}\frac{\tan(z)^{2m+1}}{2m+1}\,\mathrm{d}z$$ $$=\frac{\pi}3\ln(\sqrt{3}+2)-2\sum_{m=0}^{\infty}\frac1{2m+1}\int_0^{\pi/6}\tan(z)^{2m+1}\,\mathrm{d}z.$$ What remains is for you to prove that $$\sum_{m=0}^{\infty}\frac1{2m+1}\int_0^{\pi/6}\tan(z)^{2m+1}\,\mathrm{d}z=\frac{\beta(2)}3.$$ This should be doable from well-known results.

Answer 3

Using integration by parts: \begin{align*} \int{\frac{x}{\cos(x)}dx} &= \int{x\sec(x)\ dx} \\ &= x\sec(x)\tan(x) - \int{\sec(x)\tan(x)dx} \\ &= x\sec(x)\tan(x) - \int{\cos(x)\sec^2(x)\tan(x)dx} && ...multiply \ by \cos(x)\sec(x) = 1 \\ &= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\sin(x)\tan(x)dx}]\\ &= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\frac{\sin^2(x)}{\cos(x)}dx}] \\ &= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\frac{1-\cos^2(x)}{\cos(x)}dx}] \\ &= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{[\sec(x)-\cos(x)]dx }] \\ &= x\sec(x)\tan(x) - [\tan(x)\cos(x) + [\sec(x)\tan(x)-\sin(x)]] \\ \end{align*}

After puttting limits, we get answer = $\frac{2\pi}{\sqrt{3}} - [\frac{\sqrt{3}}{2}+[2\sqrt{3} - \frac{\sqrt{3}}{2}]]$