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Calculating the centroid of a part of rosacea $r(t) = 2a\cos(2t)$, did this definite integral. $$ \int_{0}^{\pi/4}\cos^3(2t)\cos t dt $$ $$ \cos^3(2t)\cos t = \cos t\cos(2t)\cos^2(2t) = \cos t\cos(2t)\ \frac{[1 + \cos(4t)]}{2} = ?? $$ I tried to use the technique of double bow, but could not. Can someone help me?

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    $\begingroup$ Let $u = \cos t$ and then use one of the identities to represent $\cos(2t)$ in terms of $\sin t$. $\endgroup$ – trang1618 Sep 23 '16 at 22:29
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We know: $\cos(2t) = 1-2\sin^2t$. Let $u = \sin t$, we have: $$I = \int_0^{\sqrt{2}/2}(1-2u^2)^3du$$ Then expand and integrate using power rule.

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