Answer
$$-{8\over 3\pi}$$
Work Step by Step
let $u = 1+sin t$
$$ h_{ave} = {1\over 3\pi/2 - \pi/2} \int ^{3\pi/2} _{\pi/2} (1+sin t)^2 cos t dt = {1\over \pi} \int ^0 _2 u^2 du $$
$${1\over \pi} [{u^3 \over 3}] ^0 _2 = {1\over \pi} (-{8\over 3}) = -{8\over 3\pi}$$
