$\beta$ -day Problem 2

We will use the following $B(x,y)=\int^1_0t^{x-1}(1-t)^{y-1}\,dt$ $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$ The integral $I=\int^1_0\frac{1}{\sqrt[3]{x^2(1-x}}\,dx=\int_0^1x^{-\frac{2}{3}}(1-x)^{-\frac{1}{3}}dx$ $\Rightarrow I=\int_0^1x^{\frac{1}{3}-1}(1-x)^{\frac{2}{3}-1}\,dx$ $I=B(\frac{1}{3},\frac{2}{3})=\frac{\Gamma(\frac{1}{3}\Gamma(\frac{2}{3})}{\Gamma(1)}$ $I=\Gamma(\frac{1}{3})\Gamma(1-\frac{2}{3})=\frac{\pi}{sin(\frac{\pi}{3})}=\frac{2\pi}{\sqrt{3}}$ Hence , we get our answer as $a+b=\boxed{5}$

Same as done. Here is an upvote