Search results

don't start off programming the calculator... i did and when i did real programming i used to many global variables, not enough functions and parameter passes

ahh got it now ty guys, substitution of sin wouldn't work though, used parts

eww i made more stupid mistakes :(

fun integral: integral{dx/[x+sqrt(x^2-2)]}

x = sin@ dx= cos@ d@ sqrt(1-x^2) = sqrt(1-sin^2(@))= cos(@) integral(arcsin(sin(@))dx @^2 = sin^2x

there is a separate rule for linear factors raised to a power e.g. a/x^4+b/x^3+c/x^2+d/x can be rewriten as (a+bx+cx^2+dx^3)/x^4 they still cater for the same quantity of variables but are written differently what u expected was (ax+b)/x^2 but i rewrote this as a/x+b/x^2

Re: 回复: Re: 回复: Re: Integration Question u only put cx+d if the denominator is irreducable example (cx+d)/x^2 = c/x + d/x^2 and i put a a/x^2 and b/x so it works out

Re: 回复: Re: Integration Question whats wrong with it? neat@ treblas method

i've seen people do better, a guy in my school didn't do a single past paper and got 96 in accelerated maths, another never did homework or study, hadn't even learnt all the topics and got a low band 6

use partial fractions; i don't have latex 1/((x^2)(1+x^2) = a/x^2+b/x+(cx+d)/(1+x^2) a= 1/(1+0) = 1 [heaviside cover method] (1+x^2) +bx(1+x^2) + (cx+d)x^2 = 1 substitute mutual value: x=i (1+i^2) + bi(1+i^2) + (ci+d)(i^2) = 1 (i^2=-1 so 1+i^2 = 0) ci+d = -1 c = 0, d = -1 (equating complex and...

dam that was stupid of me lol, if i took out a>0 and b>0 it would be right though

a number, this concept is usually asked in a question like "explain why integral(0, 2) x^2-1 dx will not find the area of this curve bounded by the x axis

b) from 1: a^2+b^2>=2ab, similarly b^2+c^2>=2cb and c^2+a^2>=2ac add equations 1,2 and 3 : 2(a^2+b^2+c^2) >= 2ab+2bc+2ac a^2+b^2+c^2>=ab+bc+ac

i haven't learnt harder 3u yet but i think question 1 goes like (a-b)^2=a^2-2ab+b^2 so a^2+b^2=(a+b)^2+2ab.. since a>0 and b>0 : (a+b)^2>0 hence a^2+b^2 >= 2ab

i'm bad at typing so i'll do the 1st partial fraction S(dx/[x^5-x^3]) =S(dx/((x^3)(x-1)(x+1))) = a/(x^3) + b/(x^2) + c/x + d/(x-1) + e/(x+1) d = 1/2, e = 1/2, a = -1 [heavyside cover method] a(x-1)(x+1)+b(x-1)(x+1)x+c(x-1)(x+1)x^2 + dx^3(x+1) + ex^3(x-1) equating d3; 0=c+d+e -> c=-1...

its acutally much easier to solve using partial fractions

how did u get from integral[(cos(2@)/(sin^2(@)*sin(2@))] to the next step? u just substituted without integrating... i think

if its obtuse then @-90 or |90-@|

people in the "math>apprecieating beauty and elegance"forum would generally go more towards the yes

pi, i can show of my memory when i'm bored without explaining what i'm reciting