Question

>:O

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@Jpeg16

Answer 5

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

Answer 6

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

Answer 7

okay, awesome

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

The power keeps dropping by 2 each time.

Answer 12

wait

Answer 13

I lost you a bit ago

Answer 14

My bad XD

Answer 15

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

Answer 16

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

Answer 17

wait, just a second. I think I get it

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

ah, okay

Answer 25

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

Answer 26

make a little more sense maybe? :x

Answer 27

yeah, definitely

Answer 28

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

Answer 29

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

Answer 30

Yes

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

distribute the 1/6 into I4?

Answer 36

yes

Answer 37

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.

Answer 38

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

Answer 39

do we need to apply the formula one more time?

Answer 40

THE formula, haha

Answer 41

Yah let's apply it once more.

Answer 42

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.

Answer 43

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

Answer 44

So that's what the Reduction Formula tells us.

Answer 45

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

Answer 46

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

Answer 47

So what does that I_0 become?

Answer 48

cos^0=1?

Answer 49

good, and integrating it gives?

Answer 50

integrating 1*

Answer 51

x

Answer 52

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

Answer 53

ah, awesome

Answer 54

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

Answer 55

ahahahaha....stuff

Answer 56

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

Answer 57

yes, I have it all worked out

Answer 58

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

Answer 59

sin0=1 and same for 2pi

Answer 60

Woops, you're thinking cosine.

Answer 61

yes, so it's 0

Answer 62

sin=y value

Answer 63

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

Answer 64

everything is multiplied by sin, haha

Answer 65

so after all of that the answer is 0?

Answer 66

Everything expect the x, right?

Answer 67

yep

Answer 68

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

Answer 69

no more stuff

Answer 70

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

Answer 71

pi/8

Answer 72

whoaaa

Answer 73

Yay team \c:/ we finally did it!

Answer 74

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

Answer 75

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

Answer 76

angry XD lolol

Answer 77

you're the best!

Answer 78

That reduction formula sucks but is awesome too

Answer 79

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 :[

Answer 80

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