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Re: HSC 2013 4U Marathon Ah yes, I'm making too many mistakes at the moment :/ However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?

Re: HSC 2013 4U Marathon Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =) ================== $Find$ I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx

Re: HSC 2013 4U Marathon Note the binomial co-efficient at the front =)

Re: HSC 2013 4U Marathon x^2=u du= 2x dx I = 2 \int \frac{x}{x+1} I = 2(x+1) - 2\ln(x+1) + c I = \ln \left(\frac{e^{\sqrt{u}+1}}{\sqrt{u}+1} \right )^2 + c ============================== $Prove that$ \sum_{r=0}^{n} \binom{n}{r} \cos(\alpha+2r\beta) = 2^n \cos^n \beta

Re: HSC 2013 4U Marathon $Prove the following$ $i)$ \int_0^{\pi/4} \sin(2x) \ln(\cos x) \ dx = \frac{1}{4}(\ln2-1) $ii)$ \int_0^{\pi/4} \cos(2x) \ln(cos x) \ dx = \frac{1}{8}(\pi - \ln 4 - 2) $Hence evaluate$ $iii)$ \int_{\pi/4}^{\pi/2} (\cos(2x)+\sin(2x)) \ln(\cos x + \sin x) \ dx

Re: HSC 2013 4U Marathon x\neq 0 x \geq -1/2 \frac{2x^2}{(1+x)-\sqrt{1+2x}} \leq 2x+9 \frac{2x^2(1+x+\sqrt{1+2x})}{x^2} \leq 2x+9 2+2x+\sqrt{1+2x} \leq 2x+7 \sqrt{1+2x} \leq \frac{7}{2} x\leq \frac{45}{8} Therefore the solution: \frac{-1}{2} \leq x \leq \frac{45}{8} \ , \ \ \...

Re: HSC 2013 4U Marathon u=x^2 du=2x \ dx yes I am, thank you. ========== Sean, is saying that in the interval [a,b] it is differentiable sufficient?

Re: HSC 2013 4U Marathon $Take the function$ \ \ y=f(x) $continuously differentiable in the interval$ \ \ a < x < b $and continuous in the interval$ \ \ a \leq x \leq b $Let$ \ \ A(a,f(a)) \ \ \ B(b, f(b)) $i) By considering integration, prove that the length of the arc $along the...

Re: HSC 2013 4U Marathon I=\int \frac{x^3e^x}{(x^2+1)^2} \ dx u=x^2 \Rightarrow \ \ I = \frac{1}{2}\int \frac{u e^u}{(u+1)^2} \ du = \frac{1}{2}\int e^u \cdot \left(\frac{1}{u+1} - \frac{1}{(u+1)^2} \right ) \ du $Note that$ \frac{d}{dt}f(t) e^{t} = (f(t)+f'(t))e^t $From the above...

I guess your teacher probably won't do anything for Harder 3U with you. We only have Mechanics and Harder 3U left pretty much (and a whole term to do it)

For Question 10. Think about how you would curl the sector into a cone, you will notice that the arc length of the sector, is the circumference of the base circle of the cone. Surface Area will be the area of the sector, I don't think it will include the inside surface, but if it is, multiply...

Re: HSC 2013 3U Marathon Thread $A triangle has side lengths$ \ \ a, b, c $By considering angles of the triangle or otherwise, prove that, the area of the triangle$ \ \ A \ \ $is given by$ A = \sqrt{s(s-a)(s-b)(s-c)} \ \ \ $where$ \ \ s =\frac{1}{2}(a+b+c)

Re: HSC 2013 4U Marathon $Find$ \int \frac{dx}{(x+2)\sqrt{x^2+4x-5}}

Re: HSC 2013 4U Marathon $Consider the tetrahedron shown in the diagram$ $With side lengths$ \ \ OA = a \ \ OB = b \ \ OC = c $Moreover, the perpendicular from$ \ \ O \ \ $to$ \ \ \triangle ABC \ \ $has length$ \ \ d $By considering the volume of the tetrahedron$ $Prove that$...

Re: HSC 2013 4U Marathon I= \int \frac{ \sin2x}{(1+\sin^2x)^2} \ dx $let$ \ \ u =\sin^2 x du=\sin(2x) \ dx \therefore \ \ I = \int \frac{du}{(1+u)^2} = \frac{-1}{1+u} + c \therefore \ \ I = \frac{-1}{1+\sin^2 x} + c

Re: HSC 2013 4U Marathon Yep something like that, nice work.

Re: HSC 2013 4U Marathon $Evaluate$ I= \int_0^1 x^3 \tan^{-1} \left (\frac{1-x}{1+x} \right ) \ dx

Re: HSC 2013 4U Marathon Nah definitely t=tan(x/2) I intended the person to go backwards with it, i.e. notice that the denominator is actually (t^2+1)^2 + 8t^2 With some manipulation thereon, it returns that, the integrand: \frac{1-t^2}{1+10t^2+t^4} = \frac{\frac{1-t^2}{1+t^2} \cdot...

Re: HSC 2013 4U Marathon I don't think so, but I think its a bit unreasonable of me to ask it like that with no hint whatsoever. Redone: $Find, using the substitution$ \ \ t=\tan \frac{x}{2} \int \frac{1-t^2}{1+10t^2+t^4} \ dt

Re: HSC 2013 4U Marathon That looks right and it was a little different to my method, nice work, I think the sqrt 3 was supposed to be 3 sorry about that. ============= This is quite hard: $Find$ \int \frac{1-t^2}{1+10t^2+t^4} \ dt