Question

How do you find the integral of=
ʃ(sinx)^2 (cosx)^4 dx

Answer 1

How do you find the integral of \[\int\limits_{}^{} (sinx)^{2} (cosx)^{4} dx\]
knowing that:
\[(cosx)^{2} = \frac{ 1 }{ 2 } ( 1 + \cos2x) \]

Answer 2

and that \[(sinx)^{2} = \frac{ 1 }{ 2 } (1 - \cos2x)\]

Answer 3

and that \[(\cos2x)^{3} =\cos2x (1 - (\sin2x)^{2})\]

Answer 4

well if you want to make use of that last identity I think that you should start by changing sin^2x to 1-cos^2x

Answer 5

it will still be a dreadful mess though

Answer 6

`Even Powers` of cosine and sine are such a pain.
I much prefer to use the Cosine Reduction Formula:\[\Large\bf\sf \int\limits \cos^nx\;dx =\frac{1}{n}\cos^{n-1}x \sin x+\frac{n-1}{n}\int\limits \cos^{n-2}x\;dx\]
It reduces the power on cosine by 2 each time ^

But if you prefer Using the Half-Angle Identity over and over to get rid of those squares, that will work also.

Answer 7

Can you show me the fastest one in a detailed way so i can understand?

Answer 8

oh I think I have an easier way...
let me make sure it pans out

Answer 9

okk

Answer 10

nope still a mess :P

Answer 11

xD

Answer 12

What do I do then? ç.ç

Answer 13

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

Let,\[\Large\bf I_n\quad=\quad \int\limits \cos^nx\;dx\]
Then let's deal with the cos^6 term first,\[\Large\bf I_6\quad=\quad \frac{1}{6}\cos^5x \sin x+\frac{5}{6}I_4\]So that brings us here with the equation we started with:\[\Large\bf \int\limits \cos^4x\;dx -\int\limits \cos^6x\;dx\]\[\Large\bf =\quad I_4-\left(\frac{1}{6}\cos^5x \sin x+\frac{5}{6}I_4\right)\]Now both integrals are in terms of cosine^4 so we can combine them.

Answer 14

If this is too confusing, especially with my short hand notation D: You can let me know.

Answer 15

Another thing to keep in mind is.. if we're going to use this reduction formula, your teacher would probably want you to justify where it comes from :d
Which also takes a few steps, integration by parts.

Answer 16

What happened in the first passage? How did you transform it in cos^4(x) - cos^6(x)?

Answer 17

I'm sure @zepdrix 's way will work. Given the identities you posted, I think they want you to go\[\int\sin^2x\cos^4xdx=\int(1-\cos^2x)\cos^4xdx=\int\cos^4x-\cos^6xdx\\=\int[1+\cos(2x)]^2-[1+\cos(2x)]^3\]expand and yawn...

Answer 18

oh I got it xD

Answer 19

Ya if they `gave you` those identities to use, maybe you should go with it that way :)

Answer 20

@TuringTest the last passage doesn't make it zero?

Answer 21

Sorry for the ignorance but it's the first time I'm doing this...

Answer 22

let me make it a bit bigger

Answer 23

\[\int\sin^2x\cos^4xdx=\int(1-\cos^2x)\cos^4xdx=\int\cos^4x-\cos^6xdx\\=\large\int[1+\cos(2x)]^2-[1+\cos(2x)]^3dx\]

Answer 24

now that you can read the exponents clearly, do you still think we should get zero?

Answer 25

no xD

Answer 26

Thank you for being more detailed ^^

Answer 27

:)

Answer 28

\[1 + (\cos2x)^{2} + 2 (\cos2x) - 1 - (\cos2x)^{3} - 3\cos2x - 3 (\cos2x)^{2} \]

Answer 29

\[\int\limits_{}^{} -2 (\cos2x)^{2} - 2\cos2x - (\cos2x)^{3} dx\]

Answer 30

I think it's a coefficient of 1 on the cos(2x) term in the middle there

Answer 31

-1*

Answer 32

and I totally forgot to write the factors of 1/2 in the original setup, oops

Answer 33

\[\int\sin^2x\cos^4xdx=\int(1-\cos^2x)\cos^4xdx=\int\cos^4x-\cos^6xdx\\=\large\int\frac14[1+\cos(2x)]^2-\frac18[1+\cos(2x)]^3dx\]

Answer 34

you were on the right track though... lots of tedious algebra to make mistakes in; very little interesting calculus in this problem

Answer 35

ah ok. thanks

Answer 36

Um... should I do it again and show you the result?

Answer 37

if I were you I'f get the fractions out of the way because they are annoying, and factor out 1/8, giving\[\large\frac18\int2[1+\cos(2x)]^2-[1+\cos(2x)]^3dx\]but that just makes it a little less ugly

Answer 38

however you prefer. If you want to work the whole thing out on paper, or post each step, it's up to you

Answer 39

I'll post each step. Thank you for your patience xD

Answer 40

welcome :)

Answer 41

\[\frac{ 1 }{ 8 } \int\limits_{}^{} 2( 1 + 2\cos2x + (\cos2x)^{2}) - (1+(\cos2x)^{3}+3\cos2x+3(\cos2x)^{2})\]

Answer 42

yep

Answer 43

\[\frac{ 1 }{ 8 } \int\limits_{}^{} 2+4\cos2x+2(\cos2x)^{2}-1-(\cos2x)^{3}-3\cos2x-3(\cos2x)^{2}\]

Answer 44

\[\frac{ 1 }{ 8 }\int\limits_{}^{} 1-(\cos2x)^{2}-(\cos2x)^{3}+\cos2x\]

Answer 45

I agree so far

Answer 46

\[\frac{ 1 }{ 8 }\int\limits_{}^{}1-\cos2x(1-(\sin2x)^{2})-(\cos2x)^{2} + \frac{ 1 }{ 2 }\sin2x\]

Answer 47

How do i trasform \[(\cos2x)^{2}\] knowing the other formula?

Answer 48

first of all, where did \(\frac12\sin2x\) come from?

Answer 49

cos2x

Answer 50

why 1/2 then?

Answer 51

I don't know, it's a formula I just took from my book

Answer 52

books are nice for reference, but when they don't agree with your brain you should decide which one is right
let's go back to where we agreed:\[\frac{ 1 }{ 8 }\int 1-(\cos2x)^{2}-(\cos2x)^{3}+\cos2xdx\]

Answer 53

now we use the two formulas you were given at the beginning of the problem to sub for cos^2(2x) and cos^3(2x)

Answer 54

\[\frac{ 1 }{ 8 }\int\limits_{}^{}1-(\cos2x)^{2}-\cos2x(1-(\sin2x)^{2})+\cos2x\]

Answer 55

I don't know why I sent the same message D:

Answer 56

I didn't see it o.0
but this is good so farm but you better deal with that \(\cos^22x\) part
repeat the formula you were given to simplify cos^2

Answer 57

good so far*

Answer 58

But if I know that \[(cosx)^{2}=\frac{ 1 }{ 2 }(1+\cos2x)\]

Answer 59

how do I apply it to cos2x?

Answer 60

the exact same way. just let u=2x

Answer 61

I mean (cos2x)^2

Answer 62

are you stuck or just thinking?

Answer 63

stuck xD sorryyy

Answer 64

no worries, if you had (cos(x))^2 you would know what to do, right?

Answer 65

You said that u=2x so should i put it like this?
\[\cos(\frac{ u }{ 2 })^2= \frac{ 1 }{ 2 } (1+ cosu)\]

Answer 66

well you have the right idea, but you mixed up the two ideas
we are trying to get
\[\cos^2(2x)\]let \(u=2x\)

we then have\[\cos^2u=\frac12(1+\cos2u)\]now we just undo our substitution to get...?

Answer 67

cos\[\cos^{2}2x=\frac{ 1 }{ 2 }(1+\cos4x)\]

Answer 68

yep :)

Answer 69

\[\frac{ 1 }{ 8 }\int\limits_{}^{}1-\frac{ 1 }{ 2 }(1+\cos4x)-\cos2x(1-(\sin2x )^{2}) +\cos2x\]

Answer 70

yep, getting pretty ugly, right?
now distribute and simplify...

Answer 71

I'm sleepy xD It's 1:20 in my country
\[\frac{ 1 }{ 8 }\int\limits_{}^{}1-\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }\cos4x-\cos2x+(\cos2x)(\sin2x)^{2}+\cos2x\]

Answer 72

Well you're doing really well, then. It's only 6:30 here, but I've been up since 5am, so I'm about to call it a day after this long, painful integral

Answer 73

\[\frac{ 1 }{ 8 }\int\limits_{}^{}\frac{ 1 }{ 2}-\frac{ 1 }{ 2 }\cos4x+(\cos2x) (\sin2x)^{2}\]

Answer 74

yea you're right

Answer 75

I understood the previous partss so it's ok. Thank again for helping me :D

Answer 76

I'm willing to finish this if you are, you're almost done here

Answer 77

*you I'm sleeping xD

Answer 78

fair enough, the rest can be done with a simple u-substitution. I'm sure that will be apparent after a good night's rest. See ya around!

Answer 79

Ok! byee!! C: