`int_(pi//3)^(pi//2)(sin\ theta\ d theta)/(sqrt(1+cos\ theta`

Put `u=1+cos\ theta`, then `du=-sin\ theta\ d theta`

So

`int_(pi//3)^(pi//2)(sin\ theta\ d theta)/(sqrt(1+cos\ theta))=-int_(theta=pi//3)^(theta=pi//2)(du)/sqrtu`

`=-int_(theta=pi//3)^(theta=pi//2)u^(-1//2)du`

`=-2[u^(1//2)]_(theta=pi//3)^(theta=pi//2)`

`=-2[sqrt(1+cos theta)]_(pi//3)^(pi//2)`

`=-2[sqrt(1+0)-sqrt(1+0.5)]`

`=0.449489`