`int_(pi/6)^pi(sin(theta))d theta` Evaluate the integral.

You need to evaluate the definite integral using the fundamental theorem of calculus such that `int_a^b f(x)dx = F(b) - F(a)`

`int_(pi/6)^pi sin theta d theta = -cos theta|_(pi/6)^pi`

`int_(pi/6)^pi sin theta d theta = -cos pi + cos (pi/6)`

`int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2`

Hence, evaluating the definite integral yields

` int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2.`

You need to evaluate the definite integral using the fundamental theorem of calculus such that `int_a^b f(x)dx = F(b) - F(a)`

`int_(pi/6)^pi sin theta d theta = -cos theta|_(pi/6)^pi`

`int_(pi/6)^pi sin theta d theta = -cos pi + cos (pi/6)`

`int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2`

Hence, evaluating the definite integral yields

` int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2.`