Integrate...

- DLS

Integrate...

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- DLS

\[\LARGE \int\limits_{0}^{\frac{32 \pi}{3}}\sqrt{1+\cos2x}~dx\]

- inkyvoyd

is that cos^2 x? or cos(2x)?

- DLS

latter

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## More answers

- inkyvoyd

I'd say rewrite cos(2x) as sin^2 x-cos^2 x

- inkyvoyd

*cos^2 x-sin^2 x got the order wrong

- DLS

no..
@SithsAndGiggles

- inkyvoyd

wait cos(2x)=2cos^2 x-1 that should help you

- 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

worry first about getting the integral down before you get the abs function etc right, since the cos function is periodic

- DLS

The period is PI right?

- inkyvoyd

you said it's sqrt(1-cos(2x)) correct?

- anonymous

Break it up into the "negative" and positive parts.|dw:1378661124026:dw|

- DLS

I applied cos2x=2 cos^2x - 1

- anonymous

|dw:1378661169146:dw|

- inkyvoyd

you're missing a sqrt(2) factor

- DLS

yup,so period=pi
and yeah @inkyvoyd sry abt that!

- inkyvoyd

find the integral from pi/2 to 3pi/2

- DLS

\[\LARGE \int\limits\limits\limits_{0}^{\frac{32 \pi}{3}}\sqrt{2} \left| cosx \right| dx\]

- inkyvoyd

or just from 0 to pi/2

- DLS

in the solution its from 0 to 10 pi...or the other case..0 to 11pi..why so?

- inkyvoyd

oh, the integral of cos x= sin x. do that first, and split it up into separate chunks

- 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

|dw:1378661710405:dw|

- inkyvoyd

keep adding them until you reach 10.5pi

- DLS

\[\LARGE \frac{32}{3}\int\limits\limits\limits\limits\limits_{0}^{\ \pi}\sqrt{2} \left| cosx \right| dx\]

- inkyvoyd

I believe so.

- DLS

umm cant we do it like this??

- inkyvoyd

check your answer with wolfram alpha

- DLS

........Standard computation time exceeded...
:P

- DLS

@ash2326

- 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

|dw:1378699178215:dw|

- DLS

@inkyvoyd i know the answer,its correct

- anonymous

I don't think you can just pull out that \(32/3\) in all cases, but maybe it works this time.

- DLS

what is wrong with it?

- anonymous

You'd need to be carefull around boundary cases.

- 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

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

\[
\int_0^{10\pi}f(x)\;dx+ \int_0^{2\pi/3}f(x)\;dx
\]

- ash2326

yes, you're right. There is a property like that. Good :) @DLS

- DLS

then what is the need to break it like 0 to 11 pi and 11 pi to 32pi/3 ??

- anonymous

That property will work on the \( [0,10\pi] \) interval.

- DLS

why so? :O

- anonymous

Because it is evenly divided by \(\pi/2\).

- anonymous

However \([30\pi/3,2\pi/3]\) is not evenly divided.

- DLS

not getting it..

- anonymous

I would agree with wio

- anonymous

|dw:1378700075881:dw|

- anonymous

I guess you can only use that property when n is integer

- anonymous

|dw:1378700092325:dw|

- anonymous

since f(x) is a curve in our case

- anonymous

|dw:1378700145530:dw|

- anonymous

@DLS Do you still need help? Btw, this is a fairly straight forward question =].

- 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

@genius12 yes :/

- anonymous

DLS, do you understand what I'm saying?

- DLS

u r not able to explain it better :/ or I'm confused too much lol

- anonymous

You can split up the intergral into two parts... one that is divisible by \(\pi/2\) and the remaining part.

- anonymous

\(32\pi/3\) is not divisible by \(\pi/2\).

- DLS

okay

- anonymous

\[
32\pi/3 = 64\pi/6
\]Now you want to divide by three in numerator and denominator.

- anonymous

\[
64\pi/6 = 62\pi/6+2\pi/6 = 21\pi/2+\pi/3
\]

- DLS

\[\LARGE \int\limits_{0}^{10\pi} |cosx|dx+\int\limits_{10\pi}^{10\pi+\frac{2 \pi}{3}}|cosx|dx\]

- anonymous

Actually you want to go to \(21\pi/2\). Before I cut it a bit short.

- 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

\[\LARGE \\10( \int\limits\limits\limits_{0}^{\frac{\pi}{2}} |cosx|dx+\int\limits\limits_{\frac{\pi}{2}}^{\pi}|cosx|dx)+\int\limits\limits\limits_{0}^{\frac{2\pi}{3}}|cosx|dx\]
\[\LARGE \\10( \int\limits\limits\limits_{0}^{\frac{\pi}{2}} |cosx|dx+\int\limits\limits_{\frac{\pi}{2}}^{\pi}|cosx|dx)+\int\limits\limits\limits_{0}^{\frac{\pi}{2}}|cosx|dx+\int\limits_{\frac{\pi}{2}}^{\frac{2\pi}{3}}|cosx|dx\]

- anonymous

lol, I already gave you the split.

- 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

\[\begin{split}
&\int_0^{21(\pi/2)}|\cos(x)|\;dx+\int_{21\pi/2}^{32\pi/3}|\cos(x) |\;dx\\
=&21\int_0^{\pi/2}\cos(x)\;dx+\int_{\pi/2}^{2\pi/3}-\cos(x) \;dx
\end{split}\]

- anonymous

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

Do you understand? @DLS

- DLS

yes i get it now! thanks genius12!

- anonymous

ye no problem. was wondering why this question is taking so long lol..

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