We will first use a substitution {uduβ=cosy=βsinydyβ and use the fact that {sin2y=1βcos2ycos2y=2cos2yβ1β, which transforms the integral :
β«0Ο/2βsin3ycos(2y)dy=ββ«10β(1βu2)(2u2β1)du
This integral can be easily calculated (we have used the minus sign before the integral to transform it into β«01β ) :
β«01β(3u2β2u4β1)du=[u3β52βu5βu]01β=1β52ββ1=β52β
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