The Student Room Group
Reply 1
I've totally got this question written down somewhere. Just need to find it and I can help you.
Reply 2
Thank you!!
rachio
Oh god, I can't even do the first question!! Help me:

Use the derivatives of cosecx and cotx to prove that

d[ln(cosecx + cotx)] /dx = -cosecx


Thank you if you can help.

let y=ln(cosecx + cotx)
let u=cosecx+cotx
y=ln u => dy/du=1/u
du/dx= -cosecxcotx-cosec^2x
dy/dx=dy/du.du/dx= (-coseccotx-cosec^2x)/(cosecx+cotx)
=-cosecx(cotx+cosecx)/(cosecx+cotx)=-cosecx
Reply 4
evariste
let y=ln(cosecx + cotx)
let u=cosecx+cotx
y=ln u => dy/du=1/u
du/dx= -cosecxcotx-cosec^2x
dy/dx=dy/du.du/dx= (-coseccotx-cosec^2x)/(cosecx+cotx)
=-cosecx(cotx+cosecx)/(cosecx+cotx)=-cosecx


Lol, I was this close |<--->| to doing it, I couldn't find the work so did it again, but you beat me to it.
Reply 5
Cheers guys!!
Reply 6
If anyone has the answers to this paper - that'd be great!!!
Reply 7
I finished the paper got 100% - just a case of finding the bastard.
Reply 8
That would be amazing!!
Reply 9
Hehe. I did this paper last year. Got so nervous in the exam hall, cos it was the first question (apparently loadsa people struggled with it). I came back to it at the end though, and found it easy (since I was relaxed etc).
Reply 10
Can anyone help with question four on the paper??

Use the substitution u= 1 + sinx and integration to prove that

the integral of sinxcosx(1+sinx)^5 dx = 1/42 (1 +sinx)^6 [6sinx - 1] +constant
Reply 11
rachio
Can anyone help with question four on the paper??

Use the substitution u= 1 + sinx and integration to prove that

the integral of sinxcosx(1+sinx)^5 dx = 1/42 (1 +sinx)^6 [6sinx - 1] +constant

This is an uber hard q.

I'm gonna try this one fresh again, but it's not easy.
Reply 13
u = sinx + 1
du/dx = cosx

This means if you sub this into your original integration you get:
int[(u - 1).du/dx.u^5] dx

This simplifies to:
int[(u - 1).u^5] du

int[u^6 - u^5] du ---> (u^7)/7 - (u^6)/6

(6u^7)/42 - (7u^6)/42

take a factor of 1/42(u^6) out ---> 1/42(u^6)[6u - 7]

sub sinx + 1 back in to your u's to give:
1/42(1 + sinx)^6.[6(sinx + 1) - 7]

This gives the final answer of:
1/42(1 + sinx)^6.[6sinx - 1]

Which I believe is what you need.

Hope that helps.
Reply 14

Lol, you bitch. I'd just finished writing out that beast when I read that thread :frown: :frown: :frown:
Reply 15
has any1 done question 6 on the paper. ive done part (a) but im having loads of trouble with part (b)
Reply 16
post the question please!