DLS
  • DLS
Integrate...
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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DLS
  • DLS
\[\LARGE \int\limits_{0}^{\frac{32 \pi}{3}}\sqrt{1+\cos2x}~dx\]
inkyvoyd
  • inkyvoyd
is that cos^2 x? or cos(2x)?
DLS
  • DLS
latter

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inkyvoyd
  • inkyvoyd
I'd say rewrite cos(2x) as sin^2 x-cos^2 x
inkyvoyd
  • inkyvoyd
*cos^2 x-sin^2 x got the order wrong
DLS
  • DLS
no.. @SithsAndGiggles
inkyvoyd
  • inkyvoyd
wait cos(2x)=2cos^2 x-1 that should help you
DLS
  • DLS
\[\LARGE \int\limits\limits_{0}^{\frac{32 \pi}{3}}\sqrt{\cos^2x}~dx=\LARGE \int\limits\limits_{0}^{\frac{32 \pi}{3}}\|cosx|~dx\]
inkyvoyd
  • inkyvoyd
worry first about getting the integral down before you get the abs function etc right, since the cos function is periodic
DLS
  • DLS
The period is PI right?
inkyvoyd
  • inkyvoyd
you said it's sqrt(1-cos(2x)) correct?
anonymous
  • anonymous
Break it up into the "negative" and positive parts.|dw:1378661124026:dw|
DLS
  • DLS
I applied cos2x=2 cos^2x - 1
anonymous
  • anonymous
|dw:1378661169146:dw|
inkyvoyd
  • inkyvoyd
you're missing a sqrt(2) factor
DLS
  • DLS
yup,so period=pi and yeah @inkyvoyd sry abt that!
inkyvoyd
  • inkyvoyd
find the integral from pi/2 to 3pi/2
DLS
  • DLS
\[\LARGE \int\limits\limits\limits_{0}^{\frac{32 \pi}{3}}\sqrt{2} \left| cosx \right| dx\]
inkyvoyd
  • inkyvoyd
or just from 0 to pi/2
DLS
  • DLS
in the solution its from 0 to 10 pi...or the other case..0 to 11pi..why so?
inkyvoyd
  • inkyvoyd
oh, the integral of cos x= sin x. do that first, and split it up into separate chunks
inkyvoyd
  • inkyvoyd
and take a single sine wave "hill" and simply multiply that till you reach 10.5pi; for that last 1/6 pi (I think it is) simply add in sin(pi/6)
inkyvoyd
  • inkyvoyd
|dw:1378661710405:dw|
inkyvoyd
  • inkyvoyd
keep adding them until you reach 10.5pi
DLS
  • DLS
\[\LARGE \frac{32}{3}\int\limits\limits\limits\limits\limits_{0}^{\ \pi}\sqrt{2} \left| cosx \right| dx\]
inkyvoyd
  • inkyvoyd
I believe so.
DLS
  • DLS
umm cant we do it like this??
inkyvoyd
  • inkyvoyd
check your answer with wolfram alpha
DLS
  • DLS
........Standard computation time exceeded... :P
DLS
  • DLS
@ash2326
inkyvoyd
  • inkyvoyd
http://www.wolframalpha.com/input/?i=integrate&a=*C.integrate-_*Calculator.dflt-&f2=sqrt%281%2Bcos%282x%29%29&f=Integral.integrand_sqrt%281%2Bcos%282x%29%29&f3=0&f=Integral.rangestart\u005f0&f4=32pi%2F3&f=Integral.rangeend_32pi%2F3&a=*FVarOpt.1-_**-.***Integral.variable---.**Integral.rangestart-.*Integral.rangeend---
anonymous
  • anonymous
|dw:1378699178215:dw|
DLS
  • DLS
@inkyvoyd i know the answer,its correct
anonymous
  • anonymous
I don't think you can just pull out that \(32/3\) in all cases, but maybe it works this time.
DLS
  • DLS
what is wrong with it?
anonymous
  • anonymous
You'd need to be carefull around boundary cases.
DLS
  • DLS
I think there is a property.. \[\LARGE \int\limits_{0}^{nT}f(x)dx=n \int\limits_{0}^{T}f(x)dx\] where T is the period..
ash2326
  • ash2326
Yes, you can't do that @DLS However you can change the limit using this. \[u= \frac{3}{32} x\] I don't see any helpfulness
anonymous
  • anonymous
\[ \int_0^{10\pi}f(x)\;dx+ \int_0^{2\pi/3}f(x)\;dx \]
ash2326
  • ash2326
yes, you're right. There is a property like that. Good :) @DLS
DLS
  • DLS
then what is the need to break it like 0 to 11 pi and 11 pi to 32pi/3 ??
anonymous
  • anonymous
That property will work on the \( [0,10\pi] \) interval.
DLS
  • DLS
why so? :O
anonymous
  • anonymous
Because it is evenly divided by \(\pi/2\).
anonymous
  • anonymous
However \([30\pi/3,2\pi/3]\) is not evenly divided.
DLS
  • DLS
not getting it..
anonymous
  • anonymous
I would agree with wio
anonymous
  • anonymous
|dw:1378700075881:dw|
anonymous
  • anonymous
I guess you can only use that property when n is integer
anonymous
  • anonymous
|dw:1378700092325:dw|
anonymous
  • anonymous
since f(x) is a curve in our case
anonymous
  • anonymous
|dw:1378700145530:dw|
anonymous
  • anonymous
@DLS Do you still need help? Btw, this is a fairly straight forward question =].
anonymous
  • anonymous
What you can do is \[\large \int _0^{21(\pi/2)}f(x)\;dx+\int_{21(\pi/2)}^{32\pi/2}f(x)\;dx \]
DLS
  • DLS
@genius12 yes :/
anonymous
  • anonymous
DLS, do you understand what I'm saying?
DLS
  • DLS
u r not able to explain it better :/ or I'm confused too much lol
anonymous
  • anonymous
You can split up the intergral into two parts... one that is divisible by \(\pi/2\) and the remaining part.
anonymous
  • anonymous
\(32\pi/3\) is not divisible by \(\pi/2\).
DLS
  • DLS
okay
anonymous
  • anonymous
\[ 32\pi/3 = 64\pi/6 \]Now you want to divide by three in numerator and denominator.
anonymous
  • anonymous
\[ 64\pi/6 = 62\pi/6+2\pi/6 = 21\pi/2+\pi/3 \]
DLS
  • DLS
\[\LARGE \int\limits_{0}^{10\pi} |cosx|dx+\int\limits_{10\pi}^{10\pi+\frac{2 \pi}{3}}|cosx|dx\]
anonymous
  • anonymous
Actually you want to go to \(21\pi/2\). Before I cut it a bit short.
DLS
  • DLS
\[\LARGE \\10( \int\limits\limits_{0}^{\frac{\pi}{2}} |cosx|dx+\int\limits_{\frac{\pi}{2}}^{\pi}|cosx|dx)+\int\limits\limits_{10\pi}^{10\pi+\frac{2 \pi}{3}}|cosx|dx\]
DLS
  • DLS
\[\LARGE \\10( \int\limits\limits\limits_{0}^{\frac{\pi}{2}} |cosx|dx+\int\limits\limits_{\frac{\pi}{2}}^{\pi}|cosx|dx)+\int\limits\limits\limits_{0}^{\frac{2\pi}{3}}|cosx|dx\] \[\LARGE \\10( \int\limits\limits\limits_{0}^{\frac{\pi}{2}} |cosx|dx+\int\limits\limits_{\frac{\pi}{2}}^{\pi}|cosx|dx)+\int\limits\limits\limits_{0}^{\frac{\pi}{2}}|cosx|dx+\int\limits_{\frac{\pi}{2}}^{\frac{2\pi}{3}}|cosx|dx\]
anonymous
  • anonymous
lol, I already gave you the split.
DLS
  • DLS
\[\large 11 \int\limits_{0}^{\frac{\pi}{2}}\cos x dx-10 \int\limits_{\frac{\pi}{2}}^{\pi}\cos x dx-\int\limits_{\frac{\pi}{2}}^{\frac{2\pi}{3}}\cos xdx\]
anonymous
  • anonymous
\[\begin{split} &\int_0^{21(\pi/2)}|\cos(x)|\;dx+\int_{21\pi/2}^{32\pi/3}|\cos(x) |\;dx\\ =&21\int_0^{\pi/2}\cos(x)\;dx+\int_{\pi/2}^{2\pi/3}-\cos(x) \;dx \end{split}\]
anonymous
  • anonymous
First note that cos(2x) = 2cos^2(x) - 1. Making this substitution yields: sqrt(1+cos(2x)) = sqrt(2cos^2(x)) = sqrt(2)*|cos(x)|. We re-write the integral as: √(2)∫|cos(x)| dx over [0, 32pi/3]. First we note that the area under the curve |cos(x)| over [0, 2pi) is the same as multiplying the area of cos(x) over [0,pi/2] multiplied by 4. Then 4∫cos(x) dx = 4 over [0, pi/2]. Now we know that 32pi/3 is the same as 5(2pi) + 2pi/3. This means that the original integral can be evaluated over [0, 10pi] first. We know the area under |cos(x)| for an interval of 2pi is 4 then the area of [0, 10pi] will be 20. Now we just need to find the area under |cos(x)| over [10pi, 10pi + 2pi/3] which is the same as the interval [0, 2pi/3]. We can evaluate this area by finding the area under the curve cos(x) on [0, pi/2] and then adding the area under the curve cos(x) over [pi/3, 2pi/3], i.e. ∫cos(x) dx over [0, pi/2] - ∫cos(x) dx over [pi/2, 2pi/3]. This gives us: 1 - (sqrt(3)/2 - 1) = 2 - sqrt(3)/2. So the total area under the curve |cos(x)| over [0, 32pi/3] is: ∫|cos(x)| dx over [0, 32pi/3] = 20 + (2-sqrt(3)/2) = 22-sqrt(3)/2. But we must now multiply both sides by sqrt(2) which gives us our final answer: 2∫|cos(x)| dx over [0, 32pi/3] = 2(22-sqrt(3)/2) ≈ 29.888, rounded off to 3rd decimal. @DLS @wio
anonymous
  • anonymous
Do you understand? @DLS
DLS
  • DLS
yes i get it now! thanks genius12!
anonymous
  • anonymous
ye no problem. was wondering why this question is taking so long lol..

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