Integration Help =] (1 Viewer)

[MONKEYjulz]

Hey guys,
Can anybody tell me how to integrate this:

sin (ln x) dx

It says use integration by part but i have no idea what to let what equal what.
Help is appreciated! Cheeerrs~

 

[Monsterman]

not sure if this is right but i think got somewhere when i did substitution with x=e^u

 

[Monsterman]

turns out that i am right.. so yea.. substitution with x=e^u then do by parts..

 

[untouchablecuz]

and repeat integration by parts once more; no need for substititution

 

[jet]

Just as a point of interest, even though it isn't really needed, I did it the way monsterman suggested, just to show the number of ways it can be performed (this probably isn't realistic for an exam situation)

 

[Top Secret]

^ I'm afraid your solution will not get full marks.

 

[untouchablecuz]

why

 

[MONKEYjulz]

It's a 4 marker btw
Thanks guys! I GET IT NOW! WOO

 

[kwabon]

^ I'm afraid your solution will not get full marks.

untouchable's and jetblack's solutions are right.

 

[jet]

I'm thinking it was a lack of the constant of Integration

 

[Top Secret]

I'm thinking it was a lack of the constant of Integration

Yea.

 

[Aerath]

Yea.

Don't quote me on this - but it's not as if you're gonna lose any marks for not having the constant, right?

 

[jet]

Looking at the marking criteria it says "Correct primitive". It's a tad ambiguous, but I would just say include it anyway, just in case the marker's in a bad mood.

 

[gurmies]

Don't quote me on this - but it's not as if you're gonna lose any marks for not having the constant, right?

I was told you lose the mark once. If you don't add it in another question, you won't get pinged.

 

[kwabon]

I'm thinking it was a lack of the constant of Integration

wat? didnt you add the constant already? or was that after? :-S

 

[jet]

I edited the post after realising, lol.
It doesn't really bother me anyway. In the back of my mind I know it's there.

 

[kwabon]

I edited the post after realising, lol.
It doesn't really bother me anyway. In the back of my mind I know it's there.

haha, kk good atleast i am not seeing things. (Y)

 

[biopia]

∫sin(lnx)
let u = lnx
du = dx/x
but x = e^u so
e^u.du = dx
∫e^u.sin(u)
Through integration by parts
I₀ = e^u.sin(u) - ∫e^u.cos(u) [I find it best at this step to realise you need a positive I₀ on your second application of integration by parts, as the negative in this line of working will make it negative, therefore 2I₀ will occur instead of 0I₀ (it's sometimes a mistake I make, lol)]
Second application:
I₁ = ∫e^u.cos(u)
I₁ = e^u.cos(u) + ∫e^u.sin(u)
Substituting:
I₀ = e^u.sin(u) - [e^u.cos(u) + ∫e^u.sin(u)] - {∫e^u.sin(u) = I₀}
I₀ = e^u.sin(u) - e^u.cos(u) - I₀
2I₀ = e^u.sin(u) - e^u.cos(u)
Substituting back for u and realising e^(lnx) = x
2I₀ = x.sin(lnx) - x.cos(lnx)
I₀ = x/2[sin(lnx) - cos(lnx)] + C

I did this way to finish on untouchablecuz' earlier work, while also doing it in a slightly different way. After checking it on wolframalpha, it is correct. I'm not sure if jetblack2007' solution will yield the same answer if you were to substitute a random value in, but I just thought I'd offer the way I would approach the question =].

 

[kwabon]

I'm not sure if jetblack2007' solution will yield the same answer if you were to substitute a random value in, but I just thought I'd offer the way I would approach the question =].

you sure about that, coz the working out seems right to me.
(Y)

 

[biopia]

Well, I didn't look through his working lol.
I know some of these integrals can have more than one answer .
Chances are, he probably is correct.
All I am saying is that I know I am correct because of the wolframalpha's validation on my working, and besides, I am sure the way I posed would be a better way to attack the question under exam circumstances.
It's probably just a different technique, that is completely viable, that yields a different solution.
I am not denying anyone lol!