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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon This is why it'd probably be nicer as a definite integral with limits chosen so as to make terms vanish after IBP.
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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $\noindent One way is to write $\cos x+ \sin x \equiv \sqrt{2}\cos \left(x -\frac{\pi}{4}\right)$ (or another equivalent sinusoid) and then use a reduction formula. I'll leave the integral for someone else.$
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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $\noindent I think Paradoxica meant that the $t$-substitution ends up making things nicer than what they started with (which isn't always the case).$
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $That function takes in an angle $x$ and returns the `equivalent angle' in the range $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, where by `equivalent angle', I mean the angle $\theta$ whose tan is the same as the tan of the angle $x$...
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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $\noindent Nice.$ $\noindent Here's a more HSC-friendly one: Find $\int \frac{1}{1 + \sin x }\text{ d}x$.$
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Well since the fraction is equal to 0, its numerator is 0 (note that its denominator is always positive, so we don't need to worry about potential 0 on the denominator). So we have $\sqrt{1-x^2}-x^2=0$ (since the...
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread For side lengths, you can let one of them be unit length (length 1) and another x, choosing them so that the sine of the correct angle is x in your diagram. The remaining side can be found via Pythagoras.
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Draw right-angled triangles to get expressions for $\cos \left(\sin^{-1} x\right)$ etc.$
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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $Well done!$ $\noindent \textbf{NEW QUESTION}$ $\noindent Find $\int \frac{1}{x^8 + 1}\text{ d}x$.$
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    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $\textbf{NEW QUESTION}$ $\noindent Find $\int \frac{\sin^2 x-4\sin x \cos x+3\cos^2 x}{\sin x + \cos x}\text{ d}x$.$
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon Ah right, I was thinking of the one with a t2 on the numerator too (the one that comes about when doing integral of √(tan x) ). That one was more tedious iirc. (Split into partial fractions etc.)
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    Purchasing HSC Papers Back

    It's still possible though that if they see both questions that are deserving of fewer marks and more marks awarded, they will choose to not award the marks in the question that should have received more marks. So I was asking what'd happen in such an instance.
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon The good old tedious integral of 1/(1+t^4) haha.
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon Here's another nice question. $\noindent Find $\lim _{n\to \infty} \int _0 ^\infty \frac{\mathrm{e}^{-t}\cos t}{\frac{1}{n} + nt^2} \text{ d}t$.$
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $First we prove this for $x>0$ (i.e. $0<x<1$, so $\cos ^{-1}x$ is an acute angle (between $0$ and $\frac{\pi}{2}$)). (The case for $x=0$ is easily seen by substitution: $\cos^{-1} 0 = \frac{\pi}{2}$, so the sine of this is $1$...
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon One can read about Feynman's commentary on differentiating under the integral in these links: • https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Popular_culture • http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf...
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon $\noindent Haha yeah, or write the integrand of the original integral as $\int _0 ^a x^\theta \text{ d}\theta$, and then switch order of integration.$
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon ln(a + 1)
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    Higher Level Integration Marathon & Questions

    Re: Extracurricular Integration Marathon Alternatively, use polar coordinates (the 'standard' approach I see that omegadot was alluding to).
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    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Paradoxica said $N$ was negative. This implies that $N-2$ is also negative, so $|N-2| = -(N-2) = 2-N$.$
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