integrate sin^2xcos^4x from 0 to 2pi

- anonymous

integrate sin^2xcos^4x from 0 to 2pi

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- schrodinger

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- zepdrix

Both even powers? :( Uh ohhhh

- anonymous

Right?

- 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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## More answers

- anonymous

That's pretty much as far as I got :/

- 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

Have you learned about the `Cosine Reduction Formula` ? :o

- anonymous

\[\cos ^{2}x=\frac{ 1+\cos2x }{ 2 }\]

- 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

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

oh yeah, we did learn that formula recently

- 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

As you noted earlier:\[\Large \color{#CC0033}{\cos^2x\quad=\quad \frac{1}{2}(1+\cos2x)}\]

- anonymous

so that takes care of the \[\cos ^{2}x\]

- 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

\[\Large \color{#3366CF}{2\sin x \cos x\quad=\quad \sin2x}\]
So what does sinxcosx= ?

- zepdrix

double angle identity*

- anonymous

changes it to sin2x/2?

- zepdrix

Yup sounds right!\[\Large \left(\color{#3366CF}{\frac{1}{2}\sin2x}\right)^2 \color{#CC0033}{\frac{1}{2}(1+\cos2x)}\]

- anonymous

okay, awesome

- anonymous

then you can plug it in the formula and bring your constants out

- 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

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

Darn we should have done the reduction method.
Looking back, I actually think it would've been easier than this :p hmm

- anonymous

hahahaha

- zepdrix

hmmmm

- anonymous

couldn't after that just use u sub?

- anonymous

u=cos2x du=-2sin2x

- 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

lame

- zepdrix

Ok back to the redux formula >:U deal with it!!

- 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

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

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

then lets go back :)

- 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

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

Let's widdle down the I4 and see if this makes sense to you.

- zepdrix

Grr stupid crash again >:c

- anonymous

then lets go back :)

- zepdrix

Ah poopers! Actually we should start on the I6.

- anonymous

why I6 and not 4? Should we start at the larger one?

- 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

\[\Large I_4-\color{royalblue}{I_6} \quad=\quad I_4-\color{royalblue}{\left(\frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\right)}\]

- zepdrix

The I's confusing you at all? :o

- anonymous

eh, it's fine

- 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

There should be a 2 in the last integral.

- 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

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

\[I=4\int\limits_{0}^{\frac{ \pi }{2}}\sin ^{2}x \cos ^{4}x dx ....(1)\]

- 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

Please tell if there any mistakes in my solution. :P

- 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

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

\[4*2I=\int\limits_{0}^{\frac{ \pi }{ 2 }}\sin ^{2}2xdx\]

- anonymous

\[8I=2\int\limits_{0}^{\frac{ \pi }{ 4 }}\sin ^{2}2x dx\]

- anonymous

\[4I=\int\limits_{0}^{\frac{ \pi }{4 }}\sin ^{2}\left\{ 2\left( \frac{ \pi }{4 }-x \right) \right\}dx .\]

- anonymous

\[4I=\int\limits_{0}^{\frac{ \pi }{4}}\cos ^{2} 2xdx ...\left( 4 \right)\]

- anonymous

adding (3) and (4)
\[8 I=\int\limits\limits_{0}^{\frac{ \pi }{ 4 }} \left( \sin ^{2}2x+\cos ^{2}2x \right)dx\]

- 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

now you can complete.

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