Integral 2

Let the given integral be $I$ . Make a substitution ( $x=\tan^3{\theta}$ , $dx=3\tan^2{\theta}\sec^2{\theta}\,d\theta$ ) as follows: $\begin{aligned} I &=\int_{0}^{\pi/2}{\left(\frac{d\theta}{{\left(\cos^3{\theta}+\sin^3{\theta}\right)}^{2/3}}\right)} \\ &=\int_{0}^{\pi/2}{\left(\frac{d\theta}{{\left(\cos^3{\theta}\cdot\left(1+\tan^3{\theta}\right)\right)}^{2/3}}\right)} \\ &=\int_{0}^{\pi/2}{\left(\frac{\sec^2{\theta}}{{\left(1+\tan^3{\theta}\right)}^{2/3}}\,d\theta\right)} \\ &=\frac{1}{3}\int_{0}^{\pi/2}{\left(\frac{3\tan^2{\theta}\sec^2{\theta}}{{\left(\tan^3{\theta}\cdot\left(1+\tan^3{\theta}\right)\right)}^{2/3}}\,d\theta\right)} \\ &=\frac{1}{3}\int_{0}^{\infty}{\left(\frac{dx}{{\left(x\left(1+x\right)\right)}^{2/3}}\right)} \end{aligned}$ Now make the substitution ( $x=\frac{u}{1-u}$ , $dx=\frac{du}{{\left(1-u\right)}^2}$ ): $\begin{aligned} I &=\frac{1}{3}\int_{0}^{\infty}{\left(\frac{dx}{{\left(x\left(1+x\right)\right)}^{2/3}}\right)} \\ &=\frac{1}{3}\int_{0}^{1}{\left(\frac{\frac{du}{{\left(1-u\right)}^2}}{{\left(\left(\frac{u}{1-u}\right)\left(1+\left(\frac{u}{1-u}\right)\right)\right)}^{2/3}}\right)} \\ &=\frac{1}{3}\int_{0}^{1}{\left(\frac{du}{{\left(u\left(1-u\right)\right)}^{2/3}}\right)} \\ &=\frac{1}{3}\int_{0}^{1}{\left(u^{-2/3}{\left(1-u\right)}^{-2/3}\,du\right)} \\ &=\frac{1}{3}\,\operatorname{B}{\left(\frac{1}{3},\frac{1}{3}\right)} \\ &=\frac{1}{3}\cdot\frac{\Gamma^2{\left(\frac{1}{3}\right)}}{\Gamma{\left(\frac{2}{3}\right)}} \end{aligned}$ Finally, we use Euler's reflection formula $\Gamma{\left(z\right)}\Gamma{\left(1-z\right)}=\frac{\pi}{\sin{\left(\pi z\right)}}$ to bring the answer to the desired form: $I=\frac{1}{3}\cdot\frac{\Gamma^2{\left(\frac{1}{3}\right)}}{\Gamma{\left(\frac{2}{3}\right)}}=\frac{1}{3}\cdot\frac{{\left(\frac{\pi}{\sin{\left(\pi/3\right)}}\right)}^2}{\Gamma^3{\left(\frac{2}{3}\right)}}=\frac{1}{3}\cdot\frac{4\pi^2/3}{\Gamma^3{\left(\frac{2}{3}\right)}}=\boxed{\frac{4\pi^2/9}{\Gamma^3{\left(2/3\right)}}}$ and the answer is $1000\cdot4+100\cdot9+10\cdot2+3=4923$ .