integration problemssss (1 Viewer)

[ToO LaZy ^*]

it's 4 - sqrt(3-x)^2
so the derivative of that would be 6-2x

but i don't know how to make it into 5-x

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*
it's 4 - sqrt(3-x)^2
so the derivative of that would be 6-2x

but i don't know how to make it into 5-x

 

[ToO LaZy ^*]

ok..done

 

[nike33]

yes keypad thats wha t i was thinking

 

[ToO LaZy ^*]

Originally posted by nike33
yes keypad thats wha t i was thinking

umm...but how?

 

[ToO LaZy ^*]

anywho...i got one last integration problem

 

[Constip8edSkunk]

Originally posted by ToO LaZy ^*
umm...but how?

d(sin^2 x)dx =2sinxcosx = sin2x

Originally posted by ToO LaZy ^*
anywho...i got one last integration problem

you can do it by parts where you integrate 1 and differentiate the log but theres probably a better way.

 

[ToO LaZy ^*]

Originally posted by Constip8edSkunk
d(sin^2 x)dx =2sinxcosx = sin2x

yeah, that's the way i used..
what way were the smart people using?..(ie nike and keypad)...

 

[Constip8edSkunk]

ok since the top is the derivative of the bottom then you can just integrate so it equals a log?

 

[ToO LaZy ^*]

Originally posted by Constip8edSkunk
ok since the top is the derivative of the bottom then you can just integrate so it equals a log?

 

[Xayma]

Originally posted by ToO LaZy ^*

d/dx [ln f(x)]=f'(x)/f(x) Hence the integral of k*f'(x)/f(x)=k*ln f(x) +C (where k is a constant and C is the constant of integration)

 

[KeypadSDM]

All you've done is let u = Sin²[x] and reduced the integral down to (2 + u)^-1

I just bypassed that step saying the thing to be integrated is of the form (Derivative of Function)/(Function)
Which integrates to Ln[Function] + C

 

[KeypadSDM]

And because I love doing integration by parts, because substitution is inferior and crap, here's the last integral by parts (i.e. the hard/long way):

Let S denote the integrate sign:

I = SLn[x² - 1]dx
u = Ln[x² - 1]
u' = 2x/(x² - 1)
v' = 1
v = x

:. I = xLn[x² - 1] - S(2x²/(x² - 1))dx
= xLn[x² - 1] - S(2 + 2/(x² - 1))dx
= xLn[x² - 1] - S(2 - 1/(x + 1) + 1/(x - 1))dx ***
= xLn[x² - 1] - 2x + Ln[x + 1] - Ln[x - 1] + C

W00t, go integration by parts, the superior method.

*** Note to readers, just think about it, you've got to know how to split up partial fractions of the form k/(x² - 1) and kx/(x² - 1) quickly, it's an easy method to learn

 

[ToO LaZy ^*]

Originally posted by Xayma
d/dx [ln f(x)]=f'(x)/f(x) Hence the integral of k*f'(x)/f(x)=k*ln f(x) +C (where k is a constant and C is the constant of integration)

Originally posted by KeypadSDM
All you've done is let u = Sin²[x] and reduced the integral down to (2 + u)^-1

I just bypassed that step saying the thing to be integrated is of the form (Derivative of Function)/(Function)
Which integrates to Ln[Function] + C

ohhh....you got me all confused...i thought you were doing the last question i posted..

 

[KeypadSDM]

Originally posted by ToO LaZy ^*
ohhh....you got me all confused...i thought you were doing the last question i posted..

I think parts is the only way to do it, the substitutions I was using got stupidly hard.

 

[ToO LaZy ^*]

Originally posted by KeypadSDM

:. I = xLn[x2 - 1] - S(2x2/(x2 - 1))dx
= xLn[x2 - 1] - S(2 + 2/(x2 - 1))dx
= xLn[x2 - 1] - S(2 - 1/(x + 1) + 1/(x - 1))dx ***

what method did you use to split the fraction?

 

[Teoh]

Umm
The first part is just isolating the two, so IF you actually put it over the denominator, you would end up with 2x²/x² -1
Although, I don't really know how to spot these kind of things yet
The second part is partial fractions
2 -=- (forgot the word...similar???) a(x-1) + b(x+1)

To get the -1 and + 1

I THINK
Feel free to ridicule for sillyness

 

[KeypadSDM]

2x²/(x² - 1)

For this step, it's really quite simple, and if you want to do well in integration REMEMBER IT!

2x²/(x² - 1)
= (2x² - 2 + 2)/(x² - 1)
= 2(x² - 1)/(x² - 1) + 2/(x² - 1)
= 2 + 2/(x² - 1)

The next bit is not so obvious, however, after you've done about 50,000,000 problems, it becomes VERY obvious.

2/(x² - 1)
= - 1/(x + 1) + 1/(x - 1)

This can be checked very easily, but the method?

Do the check, you'll see the method. Remember how the -1 multiplies with the (x - 1), and the 1 multiplies with the (x + 1).

This makes the x's cancel out when adding (opposite sign, leaves you with x - x), and the 1's combine when adding (as they have the same sign of +).

Learn when to spot symmetric combinations like this, they are expecially common in past papers where the polynomial on the bottom of the fraction contains x²ⁿ - k²ⁿ.

 

[ToO LaZy ^*]

Originally posted by KeypadSDM
2x²/(x² - 1)

For this step, it's really quite simple, and if you want to do well in integration REMEMBER IT!

2x²/(x² - 1)
= (2x² - 2 + 2)/(x² - 1)
= 2(x² - 1)/(x² - 1) + 2/(x² - 1)
= 2 + 2/(x² - 1)

ohhhhhhhhhhhhhhhhh...now i get it
when i do the 'not so obvious bit' i split the fraction using A + B....
but hopefully, it'll 'click in' once i've had enough practice.
thx champ..ur da best....literally..