$\displaystyle \int^{2\pi}_{0} ( -2cost sint + sin^{2}t ) dt $ I know how to integrate sin^{2} but not sure how to do the first term?
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$\displaystyle \sin(2t) = 2\sin(t)\cos(t)$
Or to integrate $\displaystyle \int -2sin(\theta)cos(\theta)d\theta$ make the substitution $\displaystyle u= cos(\theta)$. Then $\displaystyle du= -sin(\theta)d\theta$ so the integral becomes $\displaystyle 2\int u du$.
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