# trigonometric integral related centroid

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?

• Let $u = \cos t$ and then use one of the identities to represent $\cos(2t)$ in terms of $\sin t$. – trang1618 Sep 23 '16 at 22:29

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.