`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) :}`