zepdrix
  • zepdrix
>:O
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  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
\[\Large \int\limits \sin^2x \cos^4x\;dx \quad=\quad \int\limits \cos^4x\;dx-\int\limits \cos^6x\;dx\]
zepdrix
  • zepdrix
\[\Large = \quad I_4-I_6\]
zepdrix
  • zepdrix
\[\Large I_n=\frac{1}{n}\sin x \cos^{n-1}x+\frac{n-1}{n}\;I_{n-2}\]

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zepdrix
  • zepdrix
@Jpeg16
zepdrix
  • zepdrix
\[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\]
zepdrix
  • 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
  • anonymous
okay, awesome
zepdrix
  • zepdrix
Simplifies to,\[\Large \frac{1}{6}I_4-\frac{1}{6}\sin x \cos^5x\]
zepdrix
  • 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
  • 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
  • zepdrix
The power keeps dropping by 2 each time.
anonymous
  • anonymous
wait
anonymous
  • anonymous
I lost you a bit ago
zepdrix
  • zepdrix
My bad XD
anonymous
  • anonymous
lol, it's okay. How does I4-I6 simplify to that?
zepdrix
  • 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
  • anonymous
wait, just a second. I think I get it
anonymous
  • anonymous
I understand how you got that, just not the next step
zepdrix
  • zepdrix
Distributing the negative gives us,\[\Large I_4-\frac{1}{6}\sin x \cos^5x-\frac{5}{6}I_4\]
zepdrix
  • zepdrix
Then we want to combine these since they're like terms,\[\Large I_4-\frac{5}{6}I_4\]
zepdrix
  • zepdrix
\[\Large I_4 \quad=\quad \frac{6}{6}I_4\]
zepdrix
  • zepdrix
\[\Large \frac{6}{6}I_4-\frac{5}{6}I_4 \quad=\quad \frac{1}{5}I_4\]Right? :o
zepdrix
  • 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
  • anonymous
ah, okay
zepdrix
  • zepdrix
Woop that should be 1/6 in front, typo*
zepdrix
  • zepdrix
make a little more sense maybe? :x
anonymous
  • anonymous
yeah, definitely
zepdrix
  • zepdrix
\[\Large \frac{1}{6}I_4-\frac{1}{6}\sin x \cos^5x\]
anonymous
  • anonymous
I'm caught up now. We were going to apply the reduction formula a second time to I4 now, right?
zepdrix
  • zepdrix
Yes
zepdrix
  • 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
  • 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
  • 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
  • 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
  • anonymous
distribute the 1/6 into I4?
zepdrix
  • zepdrix
yes
zepdrix
  • 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
  • anonymous
so now we're at \[\frac{ 1 }{ 24 }sinxcos ^{3}x+\frac{ 1 }{ 8 }I _{2}-\frac{ 1 }{ 6 }sinxcos ^{5}x\]
anonymous
  • anonymous
do we need to apply the formula one more time?
anonymous
  • anonymous
THE formula, haha
zepdrix
  • zepdrix
Yah let's apply it once more.
zepdrix
  • 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
  • zepdrix
\[\Large I_2 \quad=\quad \frac{1}{2}\sin x \cos x+\frac{1}{2}I_0\]
zepdrix
  • zepdrix
So that's what the Reduction Formula tells us.
zepdrix
  • 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
  • zepdrix
Reminder:\[\Large I_0 \quad=\quad \int\limits \cos^0x\;dx\]
zepdrix
  • zepdrix
So what does that I_0 become?
anonymous
  • anonymous
cos^0=1?
zepdrix
  • zepdrix
good, and integrating it gives?
zepdrix
  • zepdrix
integrating 1*
anonymous
  • anonymous
x
zepdrix
  • 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
  • anonymous
ah, awesome
zepdrix
  • zepdrix
So what we're left with, after distributing the 1/8, is,\[\Large \frac{1}{16}x+(stuff)\]Right?
anonymous
  • anonymous
ahahahaha....stuff
zepdrix
  • zepdrix
Here is where the fun part comes in!! :x Make sure you're caught up to this point.
anonymous
  • anonymous
yes, I have it all worked out
zepdrix
  • 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
  • anonymous
sin0=1 and same for 2pi
zepdrix
  • zepdrix
Woops, you're thinking cosine.
anonymous
  • anonymous
yes, so it's 0
anonymous
  • anonymous
sin=y value
zepdrix
  • 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
  • anonymous
everything is multiplied by sin, haha
anonymous
  • anonymous
so after all of that the answer is 0?
zepdrix
  • zepdrix
Everything expect the x, right?
anonymous
  • anonymous
yep
zepdrix
  • zepdrix
\[\Large \frac{1}{16}x+\cancel{(stuff)}\]
anonymous
  • anonymous
no more stuff
zepdrix
  • zepdrix
So we're left with,\[\Large \frac{1}{16}x|_0^{2\pi}\]
anonymous
  • anonymous
pi/8
anonymous
  • anonymous
whoaaa
zepdrix
  • zepdrix
Yay team \c:/ we finally did it!
zepdrix
  • 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
  • anonymous
yeahhh, this method is more universal. Theirs was confusing me and making me angry
zepdrix
  • zepdrix
angry XD lolol
anonymous
  • anonymous
you're the best!
anonymous
  • anonymous
That reduction formula sucks but is awesome too
zepdrix
  • 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
  • anonymous
yeah, I realized that I got kind of lucky with that one, haha

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