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Re: MX2 Integration Marathon I=\int_0^{\pi/2}x\cot xdx =\int_0^{\pi/2}x(\ln\sin x)^\prime dx=-\int_0^{\pi/2}\ln\sin xdx $Let$ x=\frac{\pi }{2}-u, $we have$ I=-\int_0^{\pi/2}\ln\cos udu $Therefore$ 2I=-\int_0^{\pi/2}(\ln\sin x+\ln\cos x)dx =\frac\pi2\ln2-\int_0^{\pi/2}\ln\sin2xdx $I just...

Re: MX2 Integration Marathon Looks like a hard integral. I did IBP and got \frac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} cosec^{2}x dx which doesn't really help.

Re: MX2 Integration Marathon $Assuming$ a^{2}<x^{2}: \int_0^a \frac{dx}{x+\sqrt{x^2-a^2}}\times \frac{x-\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}} =\frac{1}{a^{2}}\int_0^a x-\frac{1}{a^{2}}\int_0^a\sqrt{x^2-a^2}dx =\frac{x^{2}}{2a^{2}}-\frac{x}{2a^{2}}\sqrt{x^2-a^{2}}-2 ln(\sqrt{x^2-a^{2}}+x)+C

Re: MX2 Integration Marathon $Let$ I = \int_{0}^{1} cot^{-1}(x)dx $Let$ u=cot^{-1}x, du=\frac{-1}{1+x^{2}}dx, dv=dx, v=x \therefore I = \frac{\pi }{4}+ln(\sqrt{2})

If you can't do it straight away just use a substitution. Let u=3-2x and let u=x-2.

haha same here :( It's scary how much harder maths gets.

Re: MX2 Integration Marathon Nice question. Looks hard but can be easily done using a substitution.

Hey, you seem pretty good at maths. Do you have an account on http://math.stackexchange.com/ ? Best maths forum ever!

Re: MX2 Integration Marathon Yeah page 14. No one actually did it though but people mentioned the two methods.

Re: MX2 Integration Marathon Not sure if this is correct because it looks really ugly... $Let$ I=\int \frac{sin^{-1}(e^{x})}{e^{x}} Let \frac{1}{e^{x}}=u\therefore dx=-due^{x} =-\int sin^{-1}(\frac{1}{u})du Let \frac{1}{u}=v\therefore du=-\frac{dv}{v^{2}} \therefore I=\int...

Re: MX2 Integration Marathon I think this same question is on the 4U marathon thread :) The tan sub is easier and more obvious, isn't it?

Re: HSC 2013 4U Marathon Good job! By the way, \int_{0}^{\infty }sin(x^{2})dx=\frac{1}{\sqrt{\pi }}\int_{0}^{\infty }\frac{1}{x^{4}+1}dx=\frac{\sqrt{\pi} }{2\sqrt{2}} No problem, I will post there next time.

Re: HSC 2013 4U Marathon The final result is nice though. Allows you to evaluate a sine Fresnel Integral without using complex analysis (you need Gamma function though).

Re: HSC 2013 4U Marathon This can be so easily done using abstract algebra but I don't think that is HSC level.

Re: HSC 2013 4U Marathon You can't use that substitution because you can't get rid off the x after you sub dx back. The only substitution you can use is x=tanx; \int_{0}^{\infty }\frac{sec^{2}\theta }{tan^{4}\theta+1}d\theta which doesn't really help.

Re: HSC 2013 4U Marathon $Evaluate$ \int_{0}^{\infty }\frac{1}{x^{4}+1}dx