My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree1Thanks
  • 1 Post By skipjack
LinkBack Thread Tools Display Modes
December 21st, 2015, 07:40 AM   #1
Joined: Dec 2015
From: malaysia

Posts: 66
Thanks: 1

integrate sin2x cos4x

Why the final answer that I get is = 0? Which part of my working is wrong?
Attached Images
File Type: jpg 9773.jpg (93.9 KB, 11 views)

Last edited by skipjack; December 21st, 2015 at 12:31 PM.
xl5899 is offline  
December 21st, 2015, 08:10 AM   #2
Math Team
Joined: Jul 2011
From: Texas

Posts: 2,761
Thanks: 1416

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 ...

$u = \cos(2x)$

$du=-2\sin(2x) \, dx$

$\displaystyle \int_{-1}^1 u^2 \, du - \int_0^{\pi/2} \sin(2x) \, dx$

$\bigg[\dfrac{u^3}{3} \bigg]_{-1}^1 - \bigg[-\dfrac{\cos(2x)}{2}\bigg]_0^{\pi/2}$

$\dfrac{2}{3} - 1 = -\dfrac{1}{3}$
skeeter is offline  
December 21st, 2015, 08:17 AM   #3
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,344
Thanks: 2466

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.
v8archie is offline  
December 21st, 2015, 01:04 PM   #4
Global Moderator
Joined: Dec 2006

Posts: 19,294
Thanks: 1686

$\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
Thanks from greg1313
skipjack is offline  
December 22nd, 2015, 04:46 AM   #5
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

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 09:01 AM.
Country Boy is offline  

  My Math Forum > College Math Forum > Calculus

cos4x, integrate, sin2x

Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
sin2x= square root 3/2?? How to solve? magnanimous Trigonometry 3 December 16th, 2015 11:51 AM
min value y= sin2x+cosx mared Pre-Calculus 1 May 7th, 2014 12:25 PM
integrate (cos(x/10)^3) dx Singularity Calculus 0 February 14th, 2013 04:29 PM
How do I integrate this? OriaG Calculus 4 February 2nd, 2013 02:17 PM
integral of sin2x/2sincos^2x nephi39 Calculus 6 January 1st, 2012 05:34 PM

Copyright © 2018 My Math Forum. All rights reserved.