>:O

- zepdrix

>:O

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- schrodinger

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- zepdrix

\[\Large \int\limits \sin^2x \cos^4x\;dx \quad=\quad \int\limits \cos^4x\;dx-\int\limits \cos^6x\;dx\]

- zepdrix

\[\Large = \quad I_4-I_6\]

- zepdrix

\[\Large I_n=\frac{1}{n}\sin x \cos^{n-1}x+\frac{n-1}{n}\;I_{n-2}\]

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## More answers

- zepdrix

@Jpeg16

- zepdrix

\[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\]

- zepdrix

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

- anonymous

okay, awesome

- zepdrix

Simplifies to,\[\Large \frac{1}{6}I_4-\frac{1}{6}\sin x \cos^5x\]

- zepdrix

So we need to widdle this one down a bit further,\[\Large I_4 \quad=\quad \frac{1}{4}\sin x \cos^3x+\frac{3}{4}I_2\]

- zepdrix

Understand what we're doing?
\[\Large I_4 \quad=\quad \int\limits \cos^4x\;dx\]
So we're starting with an integral of cos^4, and ending up with (stuff) + integral of cos^2

- zepdrix

The power keeps dropping by 2 each time.

- anonymous

wait

- anonymous

I lost you a bit ago

- zepdrix

My bad XD

- anonymous

lol, it's okay. How does I4-I6 simplify to that?

- zepdrix

So we need to clarify what happens after this step?
Did you understand the step that led to this setup, where I6 turned into the bracket portion?
\[\Large I_4-I_6 \quad=\quad I_4-\left(\frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\right)\]

- anonymous

wait, just a second. I think I get it

- anonymous

I understand how you got that, just not the next step

- zepdrix

Distributing the negative gives us,\[\Large I_4-\frac{1}{6}\sin x \cos^5x-\frac{5}{6}I_4\]

- zepdrix

Then we want to combine these since they're like terms,\[\Large I_4-\frac{5}{6}I_4\]

- zepdrix

\[\Large I_4 \quad=\quad \frac{6}{6}I_4\]

- zepdrix

\[\Large \frac{6}{6}I_4-\frac{5}{6}I_4 \quad=\quad \frac{1}{5}I_4\]Right? :o

- zepdrix

I think these I's are throwing you off a little bit lol :)
It's just some variable thing.
So if we (an apple) and we subtract from it (5/6 of an apple), what are we left with? :3

- anonymous

ah, okay

- zepdrix

Woop that should be 1/6 in front, typo*

- zepdrix

make a little more sense maybe? :x

- anonymous

yeah, definitely

- zepdrix

\[\Large \frac{1}{6}I_4-\frac{1}{6}\sin x \cos^5x\]

- anonymous

I'm caught up now. We were going to apply the reduction formula a second time to I4 now, right?

- zepdrix

Yes

- zepdrix

This worked out kind of nice since we were able to combine the two terms.
Widdling the 6 to 4 to 2 to 0 AND the 4 to 2 to 0 would have been a lot of work lol

- zepdrix

Applying the Reduction Formula to the 4th power integral gives us:\[\Large I_4 \quad=\quad \frac{1}{4}\sin x \cos^3x+\frac{3}{4}I_2\]

- zepdrix

\[\Large \frac{1}{6}\color{#662FFF}{\left(\frac{1}{4}\sin x \cos^3x+\frac{3}{4}I_2\right)}-\frac{1}{6}\sin x \cos^5x\]

- zepdrix

Distributing the 1/6,\[\Large \frac{3}{24}I_2+\frac{1}{24}\sin x \cos^3 x-\frac{1}{6}\sin x \cos^5x\]

- anonymous

distribute the 1/6 into I4?

- zepdrix

yes

- zepdrix

We're almost done, I know it looks like it's going to be difficult.
But after we widdle down our I2 to a I0, it's going to get very very easy for us.

- anonymous

so now we're at \[\frac{ 1 }{ 24 }sinxcos ^{3}x+\frac{ 1 }{ 8 }I _{2}-\frac{ 1 }{ 6 }sinxcos ^{5}x\]

- anonymous

do we need to apply the formula one more time?

- anonymous

THE formula, haha

- zepdrix

Yah let's apply it once more.

- zepdrix

We COULD use the Cosine Double Angle from this point instead.
But I want you to see something cool, so we should use the formula once more.

- zepdrix

\[\Large I_2 \quad=\quad \frac{1}{2}\sin x \cos x+\frac{1}{2}I_0\]

- zepdrix

So that's what the Reduction Formula tells us.

- zepdrix

Plugging in:\[\large \frac{1}{8}\left(\frac{1}{2}\sin x \cos x+\frac{1}{2}I_0\right)+\frac{1}{24}\sin x \cos^3 x-\frac{1}{6}\sin x \cos^5x\]

- zepdrix

Reminder:\[\Large I_0 \quad=\quad \int\limits \cos^0x\;dx\]

- zepdrix

So what does that I_0 become?

- anonymous

cos^0=1?

- zepdrix

good, and integrating it gives?

- zepdrix

integrating 1*

- anonymous

x

- zepdrix

Good good good,\[\large \frac{1}{8}\left(\frac{1}{2}\sin x \cos x+\frac{1}{2}x\right)+\frac{1}{24}\sin x \cos^3 x-\frac{1}{6}\sin x \cos^5x\]

- anonymous

ah, awesome

- zepdrix

So what we're left with, after distributing the 1/8, is,\[\Large \frac{1}{16}x+(stuff)\]Right?

- anonymous

ahahahaha....stuff

- zepdrix

Here is where the fun part comes in!! :x
Make sure you're caught up to this point.

- anonymous

yes, I have it all worked out

- zepdrix

We need to evaluate this entire thing at x=0 and x=2pi.
Before we worry about that, think about this a moment:\[\Large \sin0=?\]\[\Large \sin2\pi=?\]

- anonymous

sin0=1 and same for 2pi

- zepdrix

Woops, you're thinking cosine.

- anonymous

yes, so it's 0

- anonymous

sin=y value

- zepdrix

Good!
Those are our upper and lower limits.
So ANYTHING being multiplied by sinx will become 0, both for x=0 and x=2pi.
See what's going to happen? Do you see the magic?? :OO

- anonymous

everything is multiplied by sin, haha

- anonymous

so after all of that the answer is 0?

- zepdrix

Everything expect the x, right?

- anonymous

yep

- zepdrix

\[\Large \frac{1}{16}x+\cancel{(stuff)}\]

- anonymous

no more stuff

- zepdrix

So we're left with,\[\Large \frac{1}{16}x|_0^{2\pi}\]

- anonymous

pi/8

- anonymous

whoaaa

- zepdrix

Yay team \c:/ we finally did it!

- zepdrix

You might want to look back at your previous thread and decide whether or not the method the other guys used was easier for you.
I can admit, this method takes a bit more time.
It feels more systematic to me though. I like it :o

- anonymous

yeahhh, this method is more universal. Theirs was confusing me and making me angry

- zepdrix

angry XD lolol

- anonymous

you're the best!

- anonymous

That reduction formula sucks but is awesome too

- zepdrix

This method might require a bit more work if our limits of integration weren't nice easy numbers.
Like if we were going from 0 to pi/4.
We'd have to plug in for aaaaall of those terms :[

- anonymous

yeah, I realized that I got kind of lucky with that one, haha

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