My Math Forum integrate sin2x cos4x

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December 21st, 2015, 08:40 AM   #1
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integrate sin2x cos4x

Why the final answer that I get is = 0? Which part of my working is wrong?
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Last edited by skipjack; December 21st, 2015 at 01:31 PM.

 December 21st, 2015, 09:10 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,701 Thanks: 1359 maybe someone else will decipher your hieroglyphics ... meanwhile, here is an easier method using a double angle identity $\displaystyle \int_0^{\pi/2} \sin(2x)\cos(4x) \, dx$ $\displaystyle \int_0^{\pi/2} \sin(2x)\bigg[2\cos^2(2x)-1\bigg] \, dx$ $\displaystyle \int_0^{\pi/2} 2\sin(2x)\cos^2(2x)-\sin(2x) \, dx$ $\displaystyle \int_0^{\pi/2} 2\sin(2x)\cos^2(2x) \, dx - \int_0^{\pi/2} \sin(2x) \, dx$ first integral ... $u = \cos(2x)$ $du=-2\sin(2x) \, dx$ $\displaystyle \int_{-1}^1 u^2 \, du - \int_0^{\pi/2} \sin(2x) \, dx$ $\bigg[\dfrac{u^3}{3} \bigg]_{-1}^1 - \bigg[-\dfrac{\cos(2x)}{2}\bigg]_0^{\pi/2}$ $\dfrac{2}{3} - 1 = -\dfrac{1}{3}$
 December 21st, 2015, 09:17 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,148 Thanks: 2386 Math Focus: Mainly analysis and algebra When integrating by parts more than once, you must keep the same 'half' of the expression as $u$ and the same 'half' as $\mathrm d v$. If you swap, you just get back to where you were originally as you did.
 December 21st, 2015, 02:04 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,593 Thanks: 1491 \displaystyle \begin{align*}\!\int_0^{\pi/2}\! \sin(2x)\cos(4x)\, dx &= \!\int_0^{\pi/2}\! \frac12\left(\sin(6x) - \sin(2x)\right)\,dx \\ &= \frac12\left[-\frac16\cos(6x) + \frac12\cos(2x)\right]_0^{\pi/2} \\ &= \frac12\left(-\frac16\cos(3\pi) + \frac12\cos(\pi) + \frac16\cos(0) -\frac12\cos(0)\right) \\ &= \frac12\left(\frac16 - \frac12 + \frac16 - \frac12\right) \\ &= -\frac13 \end{align*} Thanks from greg1313
 December 22nd, 2015, 05:46 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,957 Thanks: 800 Actually you made several errors in your calculations. You start by doing an "integration by parts", taking $\displaystyle u= \sin(2x)$ and $\displaystyle dv= \cos(4x)dx$. Then $\displaystyle du= 2 \cos(2x)dx$ and $\displaystyle v= \frac{1}{4}\sin(4x)$ so that $\displaystyle \int_0^{\pi/2} \sin(2x) \cos(4x)dx= \left[\frac{1}{4}\sin(2x)\sin(4x)\right]_0^{\pi/2}- \frac{1}{2}\int_0^{\pi/2} \cos(2x)\sin(4x)dx$. Of course, $\displaystyle \left[\frac{1}{4}\sin(2x)\sin(4x)\right]_0^{\pi/2}= 0$ so we have $\displaystyle \int_0^{\pi/2} \sin(2x) \cos(4x)dx= - \frac{1}{2}\int_0^{\pi/2} \cos(2x)\sin(4x)dx$ That so far is correct. But now you do another integration by parts, this time choosing $\displaystyle u= \sin(4x)$, $\displaystyle dv= \cos(2x) dx$ so that $\displaystyle du= 4 \cos(4x)dx$, $\displaystyle v= \frac{1}{2}\sin(2x)$. But that just reverses the calculation you did above, integrating what you got by differentiating and differentiating what you got by integrating. It should be no surprise that this gives you, not 0, but, exactly what you started with: $\displaystyle \int_0^{\pi/2} \sin(2x) \cos(4x)dx= - \frac{1}{2}\int_0^{\pi/2} \cos(2x)\sin(4x)dx$ $\displaystyle = -\frac{1}{2}\left[\frac{1}{2}\cos(4x)\sin(2x)\right]_0^{\pi/2}+\frac{1}{2}\left(\int_0^{\pi/2} \frac{1}{2}\sin(2x))(4 \cos(4x) dx)\right)$ (You appear to have forgotten the "-1/2" in front of the integral.) Again, $\displaystyle \left[\frac{1}{2}\cos(4x)\sin(2x)\right]_0^{\pi/2}= 0$ so this reduces to $\displaystyle \int_0^{\pi/2} \sin(2x) \cos(4x)dx= \int_0^{\pi/2} \sin(2x) \cos(4x)dx$ Which does not say that the integral is 0, it just does not give a value for the integral. If, doing the second integral "by parts", you chose $\displaystyle u= \cos(2x)$ and $\displaystyle dv= \cos(4x)dx$, then you would get the $\displaystyle -\frac{1}{3}$ that skeeter and skipjack give. Last edited by skipjack; December 22nd, 2015 at 10:01 AM.

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