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oh, i thought it was quite obvious reverse chain rule OR let u=cos-1x and since u is a dummy variable, you can change the variable back to x

boredofgurmies

\\\int_{-1}^{1}\dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}}\,dx=\int_{0}^{\pi}\cos(mx)\cos(nx)\, dx\\$If $m\neq n,\\=\int_{0}^{\pi}\cos(mx)\cos(nx)\, dx\\=\frac{1}{2}\int_{0}^{\pi}(\cos(m+n)x+\cos(m-n)x)\, dx\\=0\\$If $ m=n\neq 0,\\\int_{0}^{\pi}\cos(mx)\cos(nx)\, dx\\=\int_{0}^{\pi}\cos^2(mx)\...

ill try

oops

s2

$Prove that the Fibonacci sequence, defined by: $u_0=0, u_{1}=1 $ and $u_n=u_{n-1}+u_{n-2} $ has as the formula for its $n$th term:$\\\\u_n=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n note: requires strong induction

\\S=1-r^2+r^4-...\\$For $|r|<1, S $ has a limiting sum with $a=1 $ and $ R=-r^2\\\therefore S=1-r^2+r^4-...=\frac{1}{1+r^2}\\$Integrating $S $ with respect to $r $ from $0 $ to $x, $ noting that $|x|<1\\\int_{0}^{x}\frac{1}{1+r^2}dr=\int_{0}^{x}(1-r^2+r^4-...)dr\\\ \tan^{-1}x...

harder 3u tho, its not that hard, more interesting

\\1. \ \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\frac{a+b}{\sqrt{ab}}=-3\\2. \ \int \frac{\cos x}{7+\cos2x}dx=\int \frac{\cos x}{7+1-2\sin^2x}dx=\int \frac{\cos x}{8-2\sin^2x}dx=\frac{1}{2}\int \frac{1}{4-\sin^2x}d(\sin x), $ proceed with partial fractions$

i think he means: \\$If $y=u(x)v(x)w(x), \\$Show that $ \frac{dy}{dx}=u'(x)v(x)w(x)+u(x)v'(x)w(x)+u(x)v(x)w'(x)

\\\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\y=\pm \frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}} notice that the asymptotes have general form y=\pm \frac{b}{a}x thus, to find asymptotes of a hyperbola without taking limits etc, rearrange it into the following form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 shouldn't...

\\x^2-4y^2=4\\y=\pm \frac{x}{2}\sqrt{1-\frac{4}{x^2}}\\$As $x\to \pm \infty , y\to \pm \frac{x}{2}\\\therefore $ Oblique asymptotes of the hyperbola are $y=\pm \frac{x}{2}

Re: 2 Unit Revising Marathon HSC '10 we're not idiots :p no solution

i've violated smarterchild one to many times

\\(a)\\\frac{dJ}{dt}=-\alpha R\\\frac{d^2 J}{dt^2}=\frac{d}{dt}(-\alpha R)=-\alpha \frac{dR}{dt}=-\alpha \beta J\\\therefore \frac{d^2 J}{dt^2}+\alpha \beta J=0\\(b)\\$If $J = A \cos (t\sqrt{\alpha\beta}) + B \sin (t\sqrt{\alpha\beta}) $ then,$\\\frac{d^2 J}{dt^2}+\alpha \beta J\\=\frac{d^2 (A...

http://www.facebook.com/group.php?gid=2218842866&ref=mf If Voldemort had gotten laid, none of this shit would have happened http://www.facebook.com/group.php?gid=12689384586&ref=mf I owe Smarter Child an apology

http://www.facebook.com/pages/I-Hate-When-I-Get-Crumbs-In-My-Bra/238997892683?ref=ts I Hate When I Get Crumbs In My Bra dont we all *sigh*

i've been to blacktown at night and i'm still breathing

western suburbs aren't THAT bad :o