Question

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

Answer 1

Both even powers? :( Uh ohhhh

Answer 2

Right?

Answer 3

\[\Large \int\limits (1-\cos^2x)\cos^4x\;dx\]

This would take a lot of work to actually integrate.
Maybe we can take advantage of some symmetry.
Lemme think :d

Answer 4

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

Answer 5

Hmm I don't see any fancy tricks to this one.
Looks like we might have to do it the long way.\[\Large \int\limits\cos^4x\;dx-\int\limits\cos^6x\;dx\]

Answer 6

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

Answer 7

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

Answer 8

\[\sin ^{2}x \cos ^{2}x \cos ^{2}x=\frac{ \left( 2\sin x \cos x \right)^{2} }{ 4 }*\frac{ 1+\cos 2x }{ 2 }\]
\[=\frac{ 1 }{ 8 }\sin ^{2}2x \left( 1+\cos 2x \right)\]
i think now you can solve.

Answer 9

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.

Answer 10

oh yeah, we did learn that formula recently

Answer 11

I like surji method better though XD Let's try to make that work.
You have to remember a few of your identities though :)
\[\Large \sin^2x \cos^2x \cos^2x \quad=\quad \left(\sin x \cos x\right)^2 \color{#CC0033}{\cos^2x}\]

Answer 12

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

Answer 13

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

Answer 14

Yes, now this blue part might be a little trickier to relate back to your double identity.\[\Large \left(\color{#3366CF}{\sin x \cos x}\right)^2 \color{#CC0033}{\frac{1}{2}(1+\cos2x)}\]

Answer 15

\[\Large \color{#3366CF}{2\sin x \cos x\quad=\quad \sin2x}\]

So what does sinxcosx= ?

Answer 16

double angle identity*

Answer 17

changes it to sin2x/2?

Answer 18

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

Answer 19

okay, awesome

Answer 20

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

Answer 21

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.

Answer 22

Make sure you square the 1/2 that is in the brackets! :O
Get something like this? :)
\[\Large \frac{1}{8}\int\limits \sin^22x(1+\cos2x)\;dx\]

Answer 23

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

Answer 24

hahahaha

Answer 25

hmmmm

Answer 26

couldn't after that just use u sub?

Answer 27

u=cos2x du=-2sin2x

Answer 28

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.

Answer 29

lame

Answer 30

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

Answer 31

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.

Answer 32

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

Answer 33

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.

Answer 34

then lets go back :)

Answer 35

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

Answer 36

So what we have is:\[\Large \int\limits\limits\cos^4x\;dx-\int\limits\limits\cos^6x\;dx\]We can think of this as:\[\Large I_4-I_6\]

Answer 37

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

Answer 38

Grr stupid crash again >:c

Answer 39

then lets go back :)

Answer 40

Ah poopers! Actually we should start on the I6.

Answer 41

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

Answer 42

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.
\[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^{6-1}x+\frac{6-1}{6}I_{6-2}\]
\[\Large I_6 \quad=\quad \frac{1}{6}\sin x \cos^5x+\frac{5}{6}I_4\]

Answer 43

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

Answer 44

The I's confusing you at all? :o

Answer 45

eh, it's fine

Answer 46

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.

Answer 47

There should be a 2 in the last integral.

Answer 48

\[letI=\int\limits_{0}^{2\pi}\sin ^{2}x \cos ^{4}xdx=2\int\limits_{0}^{\pi} \sin ^{2}x \cos ^{4}x dx\]

Answer 49

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.

Answer 50

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

Answer 51

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
\[\int_0^\pi \sin^2(2x)~dx\]

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

adding (1) and (2)
\[2I=\int\limits_{0}^{\frac{ \pi }{ 2 }} \sin ^{2}x \cos ^{2}x \left( \cos ^{2} x+\sin ^{2}x\right)dx\]

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

i have used in various steps
\[If f \left( 2a-x \right)=f \left( x \right)\]
\[then \int\limits_{0}^{2a}f \left( x \right)dx=2\int\limits_{0}^{a}f \left( x \right)dx\]

Answer 63

now you can complete.