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Re: MX2 Integration Marathon I_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n} x \ dx $Prove that$ \sum_{n=1}^{\infty} \frac{I_n}{2n-1} = \frac{\pi}{2}

Re: HSC 2013 4U Marathon Even though its integration I think it should be asked here because its not really a 'Q1 style' integral. ============= $In the following question, you may assume the results$ \int_0^{\infty} f(x) \ dx = \lim_{\alpha \to \infty} \int_0^{\alpha} f(x) \ dx \lim_{j...

Re: MX2 Integration Marathon You may have the answer but in a different form. But you have the idea, its just by parts then a substitution on the second integral. =================== \int \frac{1}{x^{n} + x} \ dx

Ah yes I have seen it before and I frequent it sometimes. Unfortunately a lot of the problems are way beyond first year uni (which I know bits and pieces of) so I can't do much besides ask my own problems sometimes hehe

Re: MX2 Integration Marathon Is it? Well I asked for the question to be done without that particular substitution because earlier in this Integration marathon someone already solved it using that substitution. Tan substitution is a bit obvious, though the definite integrals property I think...

Since the angle sum of a triangle is 180 degrees 180^{\circ} = \angle BAC + \theta + 45^{\circ} 135^{\circ} - \theta = \angle BAC Using the sine rule: \frac{AC}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sin(135^{\circ} - \theta)} Since this is Extension 1, we can expand the sine function...

Re: MX2 Integration Marathon x=\tan \theta I = \int_0^{\pi/4} ln(1+ \tan \theta) \ d\theta \int_0^{a} f(x) \ dx = \int_0^a f(a-x) \ dx \int_0^{\pi/4} \ln(1+\tan \theta) \ d\theta = \int_0^{\pi/4} \ln \frac{2}{1+ \tan \theta} \ d\theta 2I = \int_0^{\pi/4} \ln(2) I = \frac{\pi}{8} \ln...

Re: HSC 2013 4U Marathon Finding roots of unity then expressing in quadratic factors then applying partial fractions, then manipulating to get it into a suitable integral form: \frac{1}{x^4+1} = \frac{1}{4\sqrt{2}}\left( \frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1} -...

Re: HSC 2013 4U Marathon You have made a mistake when subbing dx back into the integral

Don't drop 2U maths.

ah ok then

Re: MX2 Integration Marathon The question is missing some information, but I'm assuming the facts you left out were: 1) Draw a line from P perpendicular to the x-axis, and that forms the trapezium? 2) When you mean area underneath the curve, I think you mean: A = \int_0^{x} \frac{1}{1+t^2} \...

I haven't Mechanics yet, sorry.

A 2 mark question: \int \frac{x}{\sqrt{1-4x^2}} \ dx And I put in: -\frac{1}{8} \sqrt{1-4x^2} + c by inspection, so without working out. Would I lose a mark? Because I have come across 3 mark questions like \int \frac{1}{x\ln(x)^3} \ dx where simple inspection will finish the problem, but...

Re: MX2 Integration Marathon Thank you And its actually Ibn Sina :P

Re: HSC 2013 3U Marathon Thread $Find without integrating$ \int_b^a \sqrt{a^2-x^2} \ dx a > b > 0

Re: MX2 Integration Marathon I = \int \frac{\sqrt{e^{2x}-1}}{e^x+1} \ dx e^x = \sec u e^x \ dx = \sec u \tan u \ du dx=\tan u \ du I = \int \frac{\tan^2 u}{\sec u +1} \ du I = \int \sec u -1 \ du I = \ln(\sec u + \tan u) - u \therefore \ I = \cos^{-1} (e^{-x}) +...

Re: MX2 Integration Marathon hehehe :s I'd prefer someone find a solution without that substitution because that has already been posted in this thread.

Re: MX2 Integration Marathon $Evaluate$ \int_0^1 \frac{\ln(1+x)}{1+x^2} \ dx $You may NOT use the substitution$ \ \ x=\frac{1-u}{1+u}

Re: HSC 2013 4U Marathon Fixed.