Integration by substitution

Romizzi · Onion

Integration by substitution

Jibblett · Sage

oh dear lord, I'll try have a dig out of my maths books

Martijn · Forum Troller

wow its been a while since if done this :)

im not sure if i read the question right though. is in sinx/1 plus rest or sinx/ (1+9(cosx)^2) ?

Romizzi · Onion

Yeah, its sinx/(1+9(cosx)^2) sorry

Waro · Senior Member

Been many years since i did this to but i think you may need to subsitute

u = cos (x)

then use sin(x) = cos (x) / tan (x) to eliminate one of the cos(x) denominators ?

Could be totally wrong though :)

Romizzi · Onion

The answer is -arctan.cosx

with 1/(1+x^2) integrating to arctan. But I need to show working, ie. which part to substitute.

audiem · Wielder of the sunglasses

1 + 1 = 2

Romizzi · Onion

Just b een playing with it in my maths software again and got a differant answer!

-1/3 arctan 3 cos x

incase that makes more sense to anyone :P

Martijn · Forum Troller

It does actually :) now the factor 9 before the (cosx)^2 is explained better :)

Hmm wait a moment. Arent you forgetting some ('s? otherwise the answer is the same 1/3*3=1 :grin:

-- Edited by Martijn at 19:15, 2008-06-30

Romizzi · Onion

-1/3 arctan.(3cos(x))

Romizzi · Onion

Just b een playing with it in my maths software again and got a differant answer!

-1/3 arctan 3 cos x

incase that makes more sense to anyone :P

Romizzi · Onion

The closest I can get it to is -1/9 arctan (9cosx)

let u = 9cosx. Then du/dx = -9sinx so constant multiple is -1/9 { in order for sinx to look like -9sinx ya have to have the multiple of -1/9 }

now integrate 1/(1+u^2) which is arctan u

re sub in u and the multiple to get -1/9 arctan (9cosx)

but Im obviously missing something or my maths software is wrong :P

BTW the formula for integ by subs is

f(g(x))g'(x) ---> f(u)du where u = g(x)

-- Edited by Romizzi at 20:05, 2008-06-30

Martijn · Forum Troller

Thats what my guess was also.
trying to find my math software to see if it gives a different answer :P

Martijn · Forum Troller

hmm 3(cosx)=u gives du/dx=-1/3sinx
then you have 1/(1+u^2)= arctan u substitute back and you have -1/3 arctan( 3cosx)

-- Edited by Martijn at 20:56, 2008-06-30

Romizzi · Onion

But Ive got 9(cosx) not 3(cosx) :P

and du/dx 3(cosx) = -3(sinx) .... no?

Martijn · Forum Troller

sorry been a bit unclear
i mean useing 3cosx=u instead of 9cosx=u

that would mean du/dx=-3sinx so you will need the factor 1/3

u^2=(3cosx)^2 -->3^2(cosx^2)=9(cosx)^2
then with the stuff you already know. (1/(1+u^2)=arctan u) you will get -1/3arctan(3cosx)
If i am right that is

Romizzi · Onion

Thats kinda makes sense, what I dont get though, is how would I know to use 3 as the multiple instead of 9, and is the 9 answer wrong?

Martijn · Forum Troller

Good question. Tbh i am not sure.
As far as i remember the subsitution is always a bit of a trying/gambling. In this case i worked to the answer you gave me, my first quess would be as you said. (sorry i cant be more convincing)

i quess you could try solveing both integrals by useing a period of 0 to pi (or 2pi) and see if they are the same.
If i recall right 3 times a cos. or 9 times a cos will be equal as a cos is a repeating wave.

But it has been quite some time since i played with this stuff so i might be wrong:grin:

Martijn · Forum Troller

thought about it a little more :P

the equation was 9*(cosx)^2 since you want to use the know equation 1/(1+u^2) you need to move the 9 "inside" the power 2. that gives you (3cosx)^2 then you subst. the 3cosx=u and the rest is know

Romizzi · Onion

Now THAT I understand! Cheers mate.

Heres another ^^ similar format, should be easy now heheh

4e^x / (1+e^2x)

I haven't looked at this myself yet

Romizzi · Onion

Here is what should be some simple maths, but as usual its the simple bits that slow me down :P
Can anyone confirm what is happening here- I cant follow the steps?

1= 7 - 3 (9 - 1 x 7)
1= -3 x 9 + 4 x 7

Jibblett · Sage

bodmas

brackets , over division, multiply, add, subtrct

so 1x7-7
9-7=2
3x2=6
7-6=1

and

4x7=28
-3x9=-27
-27+28=1

Romizzi · Onion

Ok Jibb, I understand that, but these 2 lines are related, they are steps showing the working out of something, but I dont get how they got the second line from the first one :)

Martijn · Forum Troller

work out -3(9-1*7 like -3(9-7)=-3*9+3*7
the first formula is 7-3(9-1*7) together that makes 7+3*7-3*9=4*7-3*9 aka -3*9+4*7

-- Edited by Martijn at 18:56, 2008-08-17

-- Edited by Martijn at 18:58, 2008-08-17