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k thanks. wat about int[sinx/(1+sin^2x)]?

Just wondering if there is an alternative method to t method to solve int[1/(1+sinx)]. I know t method doesnt take long but just wondering if there are alternatives. Any replies would be appreciated

you can split the (u+5)/u^0.5 into u^0.5 + 5u^-0.5 and then integrate as usual.

haha, at least you didnt end up having to do the 2nd one.

lol, its meant to be dx=-sin2@d@.

yeh i got them both now. but thanks nyway

ahh farout... stupid constant. nyway, thanks everyone

i'll try another proof for the 2nd question: let A+iB = rcos@ r>0 sqrt(A+iB) = sqrt(r)[cos(@/2)+isin(@/2)] = x+iy where x=sqrt(r)[cos(@/2)] and y=sqrt(r)[sin(@/2)] which are both real as the square root of a positive real number is real and so is the cosine and sine of a real number.

u sure? cause when you integrate RHS, you get (1+x)^(n+1)/(n+1), sub in x=-1 and it'll equal 0.

you dont have to give me the full solution. just explain how you got there. i think i'll manage.

lol, so any luck with the question? the first one seems like integration but cant seem to sub in nythin... as for the 2nd one, i got up to (1+3)^10 = 2^20 but dont no what to do with the powers of 1,3,5,7,9

need help with 2 questions: 1. By considering 1+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n=(1+x)^n, show that 1-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...+(-1)^n\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1} 2. Prove that...

sin(a+b) + sin(a-b) = 2sin(a)cos(b) let A=a+b and B=a-b sin A+sin B=2sin[(A+B)/2]cos[(A-B/2)] New Q: Factorise a^2 + 3a + 2 and hence find the coefficient of a^4 in (a^2 + 3a + 2)^6 <!-- google_ad_section_end -->

I was doing this question and i got to a point where i found that (1-2k)>0.5. I need to find an inequality with (1-2k)/k^2 in it given 0<k<0.25. Can i just go from the first step to [(1-2k)/k^2]>8 or do i need some sort of explanation. If so, what should i write?

ok i'll start from the part where you got your roots of unity as 1, cis(2pi/7), cis(-2pi/7), cis(4pi/7), cis(-4pi/7), cis(6pi/7), cis(-6pi/7). let w=cis(2pi/7) (w + w^(-1))(w^2 + w^(-2))(w^3 + w^(-3)) = (2cos(2pi/7))(2cos(4pi/7))(2cos(6pi/7)) by expanding you get...

lol yeh, i mean n is even and is bigger than or equal to 6. actually the question sucks. someone else can ask another one.

the question has alpha, not omega though

Prove by induction that a regular polygon with n sides (where n is even) has at least one side parallel to a diagonal.

if they ask a question lik "show that if @ is one of the complex roots of..., then @^2, @^3, @^4... are the other complex roots" can you just let @ be the root with smallest argument and go from there. or do you have to consider other cases?

ok i got it, but its kinda hard to explain without a diagram: In regular pentagon ABCDE, construct AD and AC and BD. BD and AC intersect at a point X. (i) φ/2 = cos(π/5) [by drawing pendicular from side to diagonal] φ = 2cos (π/5) (ii) Let AD = AC = x angle(AED)= 540/5=108...