integrating sin^2 x (1 Viewer)

[SoulSearcher]

lol sin (x*pi/90) >.>" close enough i just got lazy hahah
anyway thanks for the help

i dont need to know any of this 'integration by parts' for ext 1 do i? "

Absolutely not.

Anyway, to do it by parts, (omitting the constants)

Int. [sin²x]dx = xsin²x - Int. [x*2sinxcosx]dx

Now,
Int. [x*2sinxcosx]dx
= Int. [x*sin2x]dx
= (-xcos2x)/2 - Int. [-1/2 * cos2x]dx
= (-xcos2x)/2 + 1/2 Int. [cos2x]dx
= (-xcos2x)/2 + (sin2x)/4

Therefore,
xsin²x - Int. [x*2sinxcosx]dx
= xsin²x + (xcos2x)/2 - (sin2x)/4
= x(1 - cos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (xcos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (sin2x)/4

Therefore Int. [sin²x]dx = x/2 - (sin2x)/4 + C

 

[jemsta]

lol, and my assumptions were correct....a lot more work then just using the double angle formula
but whatever floats your boat

 

[SoulSearcher]

lol, and my assumptions were correct....a lot more work then just using the double angle formula
but whatever floats your boat

Yeah, rather tedious, but more tedious to type it all up

I was gonna post those exact same things! Damn, you're too fast

:rofl:

 

[akuchan]

Absolutely not.

Anyway, to do it by parts, (omitting the constants)

Int. [sin²x]dx = xsin²x - Int. [x*2sinxcosx]dx

Now,
Int. [x*2sinxcosx]dx
= Int. [x*sin2x]dx
= (-xcos2x)/2 - Int. [-1/2 * cos2x]dx
= (-xcos2x)/2 + 1/2 Int. [cos2x]dx
= (-xcos2x)/2 + (sin2x)/4

Therefore,
xsin²x - Int. [x*2sinxcosx]dx
= xsin²x + (xcos2x)/2 - (sin2x)/4
= x(1 - cos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (xcos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (sin2x)/4

Therefore Int. [sin²x]dx = x/2 - (sin2x)/4 + C

x_x" ahhh my eyes T.T"

 

[SoulSearcher]

x_x" ahhh my eyes T.T"

Yeah, a rather complicated solution which you could use a different method to get the same answer in a much easier way

 

[SoulSearcher]

Thats not complicated. Doing it any other way would be complicated.


Oh, <3 high levels of math.

:rofl:

I know

 

[SoulSearcher]

Im sure you find hard math questions more appealing in a different way to me...

keke

 

[jb_nc]

Do some of the SUMS problems then: http://www.maths.usyd.edu.au/u/SUMS/sums2007.pdf

 

[jemsta]

i have absolutely no idea
leave it up to the smart maths geniuses

 

[bringbackshred]

Absolutely not.

Anyway, to do it by parts, (omitting the constants)

Int. [sin²x]dx = xsin²x - Int. [x*2sinxcosx]dx

Now,
Int. [x*2sinxcosx]dx
= Int. [x*sin2x]dx
= (-xcos2x)/2 - Int. [-1/2 * cos2x]dx
= (-xcos2x)/2 + 1/2 Int. [cos2x]dx
= (-xcos2x)/2 + (sin2x)/4

Therefore,
xsin²x - Int. [x*2sinxcosx]dx
= xsin²x + (xcos2x)/2 - (sin2x)/4
= x(1 - cos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (xcos2x)/2 + (xcos2x)/2 - (sin2x)/4
= x/2 - (sin2x)/4

Therefore Int. [sin²x]dx = x/2 - (sin2x)/4 + C

NO U.

 

[SoulSearcher]

Do some of the SUMS problems then: http://www.maths.usyd.edu.au/u/SUMS/sums2007.pdf

Nope, I'm out.

 

[jb_nc]

Nope, I'm out.

I'm rather certain you can complete all or them with 4 unit knowledge considering it's open to first-years. But the people who complete them would be in the advanced and SSP programmes.

 

[SoulSearcher]

I'm rather certain you can complete all or them with 4 unit knowledge considering it's open to first-years. But the people who complete them would be in the advanced and SSP programmes.

EDIT: I just finished 5, completed it after 3 days lol.

Looking at them again, I partially understand a couple of them, and could even see where a possible solution may come from (or just deluding myself ), but I wouldn't have a clue how to actually solve them yet.

 

[bos1234]

how fast do they go thru the 4unit syllabus in uni?

 

[pLuvia]

First few weeks of uni lol. Really depends on the course you are doing though. In uni we've only covered complex numbers

 

[*chrissie*]

wat the hell's ur problem?! just cos ur jealous of our wonderfully useful maths abilities... well, i find it much easier to use substitution, (u=2x) as dy/dx is so easy to use and force into the equation.

jb_nc: ur questions look interesting . ill have a look at them... and give them to some of my awesome nerdy friends who could probably do them in 5 min... anyway, thank god we'll only have to do that sort of thing in uni!

 

[akuchan]

cant believe my simple 'integrate sin^2 x' topic lasted this long "

 

[nichhhole]

wat the hell's ur problem?! just cos ur jealous of our wonderfully useful maths abilities... well, i find it much easier to use substitution, (u=2x) as dy/dx is so easy to use and force into the equation.

edit:

misinterpreted where she was applying.
oops !

 

[gokuyt]

i du it this way
using double angle formula

from double angle formula
cos2x = 1-2sin^2x
therefore
(1-cos2x)/2 = sin^2x

the question states Int. (sin^2x) dx
substitute the value of sin^2x in the question
therefore u get
Int. (1-cos2x)/2 dx
= x/2 - (sin2x)/4 + C {cos2x = (sin2x)/2 }

 

[SoulSearcher]

i du it this way
using double angle formula

from double angle formula
cos2x = 1-2sin^2x
therefore
(1-cos2x)/2 = sin^2x

the question states Int. (sin^2x) dx
substitute the value of sin^2x in the question
therefore u get
Int. (1-cos2x)/2 dx
= x/2 - (sin2x)/4 + C {cos2x = (sin2x)/2 }

Which is the quick and easy way for integrating sin²x.