anonymous
  • anonymous
integrate sin^2xcos^4x from 0 to 2pi
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
Both even powers? :( Uh ohhhh
anonymous
  • anonymous
Right?
zepdrix
  • zepdrix
\[\Large \int\limits (1-\cos^2x)\cos^4x\;dx\] This would take a lot of work to actually integrate. Maybe we can take advantage of some symmetry. Lemme think :d

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anonymous
  • anonymous
That's pretty much as far as I got :/
zepdrix
  • zepdrix
Hmm I don't see any fancy tricks to this one. Looks like we might have to do it the long way.\[\Large \int\limits\cos^4x\;dx-\int\limits\cos^6x\;dx\]
zepdrix
  • zepdrix
Have you learned about the `Cosine Reduction Formula` ? :o
anonymous
  • anonymous
\[\cos ^{2}x=\frac{ 1+\cos2x }{ 2 }\]
anonymous
  • anonymous
\[\sin ^{2}x \cos ^{2}x \cos ^{2}x=\frac{ \left( 2\sin x \cos x \right)^{2} }{ 4 }*\frac{ 1+\cos 2x }{ 2 }\] \[=\frac{ 1 }{ 8 }\sin ^{2}2x \left( 1+\cos 2x \right)\] i think now you can solve.
zepdrix
  • zepdrix
Peg I was referring to the reduction formula for powers greater than 2:\[\Large \int\limits \cos^nx\;dx \quad=\quad \frac{1}{n}\sin x \cos^{n-1}x+\frac{n-1}{n}\int\limits \cos^{n-2}x\;dx\] But surji's method is much simpler if you're able to make sense of it.
anonymous
  • anonymous
oh yeah, we did learn that formula recently
zepdrix
  • zepdrix
I like surji method better though XD Let's try to make that work. You have to remember a few of your identities though :) \[\Large \sin^2x \cos^2x \cos^2x \quad=\quad \left(\sin x \cos x\right)^2 \color{#CC0033}{\cos^2x}\]
zepdrix
  • zepdrix
As you noted earlier:\[\Large \color{#CC0033}{\cos^2x\quad=\quad \frac{1}{2}(1+\cos2x)}\]
anonymous
  • anonymous
so that takes care of the \[\cos ^{2}x\]
zepdrix
  • zepdrix
Yes, now this blue part might be a little trickier to relate back to your double identity.\[\Large \left(\color{#3366CF}{\sin x \cos x}\right)^2 \color{#CC0033}{\frac{1}{2}(1+\cos2x)}\]
zepdrix
  • zepdrix
\[\Large \color{#3366CF}{2\sin x \cos x\quad=\quad \sin2x}\] So what does sinxcosx= ?
zepdrix
  • zepdrix
double angle identity*
anonymous
  • anonymous
changes it to sin2x/2?
zepdrix
  • zepdrix
Yup sounds right!\[\Large \left(\color{#3366CF}{\frac{1}{2}\sin2x}\right)^2 \color{#CC0033}{\frac{1}{2}(1+\cos2x)}\]
anonymous
  • anonymous
okay, awesome
anonymous
  • anonymous
then you can plug it in the formula and bring your constants out
zepdrix
  • zepdrix
We've decided to take a different route. This method won't involve using the Reduction formula. But yes let's pull the constants out first.
zepdrix
  • zepdrix
Make sure you square the 1/2 that is in the brackets! :O Get something like this? :) \[\Large \frac{1}{8}\int\limits \sin^22x(1+\cos2x)\;dx\]
zepdrix
  • zepdrix
Darn we should have done the reduction method. Looking back, I actually think it would've been easier than this :p hmm
anonymous
  • anonymous
hahahaha
zepdrix
  • zepdrix
hmmmm
anonymous
  • anonymous
couldn't after that just use u sub?
anonymous
  • anonymous
u=cos2x du=-2sin2x
zepdrix
  • zepdrix
No, since the square is on the sine term :( We'd have to split it into two interals, The second one of which we could apply a usub, But the first one we'd have to apply double angle again.
anonymous
  • anonymous
lame
zepdrix
  • zepdrix
Ok back to the redux formula >:U deal with it!!
zepdrix
  • zepdrix
I dunno, we've come so far already :p maybe it makes sense just to finish it up with the method we're on lol.
anonymous
  • anonymous
eh, whatever. If I have to go back I have to go back. At least I'll know how to do it for future problems
zepdrix
  • zepdrix
Well that's the only thing I'm worried about. The way we're doing it now will work out just fine. But the Reduction Formula will make a bit more sense than this if you have to deal with different powers. That is a neat little trick that Surji came up with, applying Double Angle Formulas over and over. But I don't think it's the way you'll want to approach these in general.
anonymous
  • anonymous
then lets go back :)
zepdrix
  • zepdrix
If we look at the Redux Formula:\[\large \int\limits\limits \cos^nx\;dx \quad=\quad \frac{1}{n}\sin x \cos^{n-1}x+\frac{n-1}{n}\int\limits\limits \cos^{n-2}x\;dx\] We're going to write it in a way that is a little bit more abbreviated.\[\Large I_n \quad=\quad \frac{1}{n}\sin x \cos^{n-1}x+\frac{n-1}{n}\;I_{n-2}\]
zepdrix
  • zepdrix
So what we have is:\[\Large \int\limits\limits\cos^4x\;dx-\int\limits\limits\cos^6x\;dx\]We can think of this as:\[\Large I_4-I_6\]
zepdrix
  • zepdrix
Let's widdle down the I4 and see if this makes sense to you.
zepdrix
  • zepdrix
Grr stupid crash again >:c
anonymous
  • anonymous
then lets go back :)
zepdrix
  • zepdrix
Ah poopers! Actually we should start on the I6.
anonymous
  • anonymous
why I6 and not 4? Should we start at the larger one?
zepdrix
  • zepdrix
What happens is, when you use this reduction formula, it reduces the power on cosine by 2 each time you do it. So after we apply this formula `1 time` it will widdle it down to something that includes cos^4x. At which point we can just combine it with the first integral thingy-ma-bobber. \[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^{6-1}x+\frac{6-1}{6}I_{6-2}\] \[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\]
zepdrix
  • zepdrix
\[\Large I_4-\color{royalblue}{I_6} \quad=\quad I_4-\color{royalblue}{\left(\frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\right)}\]
zepdrix
  • zepdrix
The I's confusing you at all? :o
anonymous
  • anonymous
eh, it's fine
blockcolder
  • blockcolder
I have a different idea: Separate the integral into 2 parts, one from 0 to pi (call this I), the other from pi to 2pi (call this J). For the first one, observe that \[\int_0^af(x)~dx=\int_0^a f(a-x)~dx\] Thus, we can see that \[I=\int_0^\pi \sin^2(x)\cos^4(x)~dx=\int_0^\pi \cos^2(x)\sin^4(x)~dx\] And \[I=2\int_0^\pi \sin^2(x)\cos^2(x)(\cos^2(x)+\sin^2(x))~dx=\int_0^\pi \sin^2(x)\cos^2(x)~dx\] so at the very least, I is simpler.
blockcolder
  • blockcolder
There should be a 2 in the last integral.
anonymous
  • anonymous
\[letI=\int\limits_{0}^{2\pi}\sin ^{2}x \cos ^{4}xdx=2\int\limits_{0}^{\pi} \sin ^{2}x \cos ^{4}x dx\]
blockcolder
  • blockcolder
You can simplify it even further by noting that sin(x) cos(x)=sin(2x)/2 so that \[I=\int_0^\pi {1 \over 2}\sin^2(2x)~dx\] which is a lot simpler than what we had.
anonymous
  • anonymous
\[I=4\int\limits_{0}^{\frac{ \pi }{2}}\sin ^{2}x \cos ^{4}x dx ....(1)\]
blockcolder
  • blockcolder
For J, make the substitution u=x-pi: \[J=\int_0^\pi \sin^2(u+\pi)\cos^4(u+\pi)~du=\int_0^\pi (-\sin(u))^2(-\cos(u))^4~du=I\] so what we did previously with I, we do it again here so that \[J=\int_0^\pi {1\over2}\sin^2(2x)~dx\] So that in the end, the whole integral (which is I+J) becomes simply \[\int_0^\pi \sin^2(2x)~dx\]
blockcolder
  • blockcolder
Please tell if there any mistakes in my solution. :P
anonymous
  • anonymous
\[I=4\int\limits_{0}^{\frac{ \pi }{ 2 }} \sin ^{2 }\left( \frac{ \pi }{2 }-x \right)\cos ^{4}\left( \frac{ \pi }{2 }-x \right)dx\]
anonymous
  • anonymous
\[I=4\int\limits_{0}^{\frac{ \pi }{ 2 }} \sin ^{2 }\left( \frac{ \pi }{2 }-x \right)\cos ^{4}\left( \frac{ \pi }{2 }-x \right)dx\]
anonymous
  • anonymous
\[or I=\int\limits_{0}^{\frac{ \pi }{2}} \cos ^{2}x \sin ^{4}x dx....(2)\]
anonymous
  • anonymous
adding (1) and (2) \[2I=\int\limits_{0}^{\frac{ \pi }{ 2 }} \sin ^{2}x \cos ^{2}x \left( \cos ^{2} x+\sin ^{2}x\right)dx\]
anonymous
  • anonymous
\[4*2I=\int\limits_{0}^{\frac{ \pi }{ 2 }}\sin ^{2}2xdx\]
anonymous
  • anonymous
\[8I=2\int\limits_{0}^{\frac{ \pi }{ 4 }}\sin ^{2}2x dx\]
anonymous
  • anonymous
\[4I=\int\limits_{0}^{\frac{ \pi }{4 }}\sin ^{2}\left\{ 2\left( \frac{ \pi }{4 }-x \right) \right\}dx .\]
anonymous
  • anonymous
\[4I=\int\limits_{0}^{\frac{ \pi }{4}}\cos ^{2} 2xdx ...\left( 4 \right)\]
anonymous
  • anonymous
adding (3) and (4) \[8 I=\int\limits\limits_{0}^{\frac{ \pi }{ 4 }} \left( \sin ^{2}2x+\cos ^{2}2x \right)dx\]
anonymous
  • anonymous
i have used in various steps \[If f \left( 2a-x \right)=f \left( x \right)\] \[then \int\limits_{0}^{2a}f \left( x \right)dx=2\int\limits_{0}^{a}f \left( x \right)dx\]
anonymous
  • anonymous
now you can complete.

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