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Re: MX2 Integration Marathon Oops, that is a typo, it should read (x+7)^5/(x+2)^7 Which can be resolved through substitution. ====== I'm running out of cool integrals =(

Re: HSC 2013 4U Marathon a, b, c \ \ $are positive real numbers$ $By considering the cases where$ \ \ a, b, c \ \ $are sides of a triangle, and when they are not sides of a triangle, prove that for all positive real values$ (-a+b+c)(a-b+c)(a+b-c) \leq abc

Re: HSC 2013 3U Marathon Thread Nice work. You can make a new lines by typing \\ i.e. [tex ]$Line 1$ \\ $Line 2$ [/ tex] ====== $Solve$ \sin^{-1}(3x+1) = \cos^{-1}(x)

Re: MX2 Integration Marathon \int \frac{(x+7)^5}{(x+2)^5} \ dx

Re: HSC 2013 3U Marathon Thread I think it would be more correct to say that the region bounded by the ln(x) curve from x=1 to x=0.5 lies below the x-axis since, \ln(x) < 0 \ \ $for$ \ \ 0<x<1 Therefore the region lies below x-axis, and hence the integral must be negative. ===== $By using...

Re: HSC 2013 4U Marathon $Without repeated iterations of Newton's Method of Approximation$ $Find a rational approximation of$ \ \ \pi \ \ $of an error less than$ \ \ 10^{-3} $Justify your answer$ EDIT: On hindsight its actually very simple, 3141/1000 will do =( ==========

Yep this is what it means. It can be proven through repeated use of the angle between 2 lines formula, first find the gradient of the tangent at some P, then gradient PS, and PS'. Find the angle between PS and the tangent, and PS' and the tangent, if they are the same, then proof is complete...

Re: HSC 2013 3U Marathon Thread Yep nice work.

Re: HSC 2013 2U Marathon $Solve for$ \ \ x \ln(1+x^2) + 1 = \ln(4+x^2)

Re: HSC 2013 3U Marathon Thread $Find the value of the sum, where$ \ \ n \ \ $is an integer$ \tan^{-1} \frac{1}{n} + \dots + \tan^{-1} \frac{1}{3} + \tan^{-1}\frac{1}{2} + \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 + \dots + \tan^{-1} n

Re: MX2 Integration Marathon yep I see, well then there you go. My method was using int f(x) = int f(a-x), then applying double angle formula to get a secant integration

Re: MX2 Integration Marathon u^2 = 1-\sin x \ \ 2u \ du = -\cos x \ dx The cos x is on the denominator, and so you cannot substitute du in easily without creating another square root.

Re: HSC 2013 4U Marathon Yep, its supposed to read fired with a velocity, at theta to the horizontal, in your final answer theta will be independent of V.

Re: MX2 Integration Marathon \int_0^{\frac{\pi}{2}} \sqrt{\frac{1}{\sin x + 1}} \ dx

Re: HSC 2013 4U Marathon Deriving the general equations of motion x= Ut \cos \theta y = -\frac{1}{2}gt^2 + Ut \sin \theta + h Their equation reminds me of the cartesian equation of motion due to the tan theta and secant squared, so finding that: t = \frac{x}{U \cos \theta} \Rightarrow...

Re: MX2 Integration Marathon That looks right, a substitution of x=u^n just resolves into a u^n ln u, so you then do by parts then you get the answer. Most probably the answers are the same. ======== \int_{-\pi}^{\pi} \sqrt{1+\cos x} \ dx

Re: MX2 Integration Marathon IBP works, but your second term is incorrect. When i made the integral I was thinking of a x=u^n substitution, turns out that it can just be done through normal IBP :s

Re: MX2 Integration Marathon Right you are, thanks.

Re: MX2 Integration Marathon By using s^2+c^2 = 1, then multiplying top and bottom by secant squared I = \int_0^{\pi /2} \frac{1}{2\sin^2 x + \cos^2 x} = \int_0^{\pi /2} \frac{\sec^2x}{\tan^2x + \sec^2 x} \ dx u = \tan x I = \int_0^{\infty} \frac{du}{2u^2+1} = \frac{1}{2} \int_0^{\infty}...

Re: MX2 Integration Marathon u^2 = \tan x 2u \ du = (1+u^4) \ dx dx = \frac{2u \ du}{1+u^4} \int \sqrt{\tan x} \ dx = \int \frac{2u^2}{1+u^4} \ du $By resolving$ a^4+1 = (a^2 + a\sqrt{2} + 1)(a^2 - a\sqrt{2} + 1) $Then a partial fraction decomposition follows, then the integral is...