integrating sin^2 x (2 Viewers)

SoulSearcher

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akuchan said:
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? o_O"
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

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

akuchan

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SoulSearcher said:
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"
 

jemsta

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

bringbackshred

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SoulSearcher said:
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.
 

jb_nc

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SoulSearcher said:
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.
 
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SoulSearcher

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jb_nc said:
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 :eek:), but I wouldn't have a clue how to actually solve them yet.
 
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pLuvia

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

*chrissie*

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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... :eek: anyway, thank god we'll only have to do that sort of thing in uni!
 

akuchan

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

nichhhole

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*chrissie* said:
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 :)!
 
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gokuyt

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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

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gokuyt said:
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.
 

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