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 December 1st, 2008, 01:02 AM #1 Newbie   Joined: Jun 2008 Posts: 28 Thanks: 0 How would you solve this integral? the integral of sin^2(u)cos(u) du (from 0 to 2pi)
 December 1st, 2008, 01:03 AM #2 Newbie   Joined: Jun 2008 Posts: 28 Thanks: 0 Re: How would you solve this integral? please help. i'm so confused. : (
 December 1st, 2008, 01:35 AM #3 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Re: How would you solve this integral? You have: $\int ^{2\pi}_0 \sin^2 u \,\cos u \,du$ Try calculating this first: $\int \sin^2 u \,\cos u \,du$ Take the substitution: $\sin u= t$ $\cos u \,du= dt \Rightarrow du = \frac{dt}{\cos u}$ Now you have: $\int \sin^2 u\, \cos u \,du= \int t^2\, \cos u\, \frac{dt}{\cos u} = \int t^2 \, dt = \frac{t^3}{3} + C = \frac{\sin^3 t}{3}$ Now the rest is simple..
 December 1st, 2008, 08:47 AM #4 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: How would you solve this integral? I'd not advise mixing variables in the same integral at the same time, nor substitution unless absolutely necessary. What you might consider is that a number of integral problems arise as a consequence of being the inverse of some already studied differential problems: $\begin{array}{l} d\sin ^3 u = 3\sin ^2 u\cos udu \\ So... \\ \int\limits_0^{2\pi } {\sin ^2 u\cos udu} = \frac{1}{3}\int\limits_0^{2\pi } {d\sin ^3 u} = \left. {\frac{{\sin ^3 u}}{3}} \right|_0^{2\pi } \\ \end{array}$ at which time you might consider that the curve is above the x-axis from 0 to Pi, and below from Pi to 2Pi.

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