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Rearrange to x^2 = -16 + 30i Let x = a + ib, expand LHS and equate real & imaginary terms/coefficients. Simultaneous equations to get values for a and b Edit: Wow, 2014 graduate? You're sure getting ahead

Re: MX2 Integration Marathon He added the definite integrals of xcot(x) and (\pi/2 -x)tan(x) together. As they're both the same, it becomes 2I = etc. He then put them both under the same denominator and multiplied numerator and denominator by 2.

Re: MX2 Integration Marathon you don't have an error, i think he forgot the /2

Good analogy, but should mention that there are more covalent bonds to break.

http://users.tpg.com.au/nanahcub/4u-draftsample.pdf i'm unsure whether questions like q7 will come up

i should really delete that, wasn't being srs

lol i know its frowned upon by teachers, but i don't know whether it's against the rules. btw, i didn't imply that it is your related text. i highly doubt you'll be studying it under belonging. you get to choose your own related texts

erm. "as you like it" is the belonging text and "blade runner" is your module a text (along with frankenstein)

10 weeks. Had me scared there..

Re: MX2 Integration Marathon so we can use other substitutions right $Let $ u = 1 + x^2 du = 2xdx \int \frac{1}{2\sqrt{u(u - 1)}}du = \int \frac{1}{2\sqrt{(u - \frac{1}{2})^2 - \frac{1}{4}}}du = \frac{1}{2}ln|u - \frac{1}{2} + \sqrt{u^2 - u}| + c = \frac{1}{2}ln|x^2 + \frac{1}{2} +...

Re: MX2 Integration Marathon $Let $ x = sin^2u dx = 2sinucosu du \int \frac{\sqrt{1 - sin^2u}}{1 - \sqrt{sin^2u}}\times 2sinucosu du = \int \frac{2sinucos^2u}{1-sinu}}du = \int \frac{2sinu(1 + sinu)(1 - sinu)}{1-sinu}du = \int 2sinu(1 + sinu)du = \int 2sinu + 1 - cos2udu = u - 2cosu -...

Re: MX2 Integration Marathon should be making your own thread. use the change sinx - cosx in to the form of Rsin(x - alpha), or you can use t substitution

Re: MX2 Integration Marathon the most recent one i posted :P spirals one on the first page i think all the ones "U MAD BRO" posted up this page

Re: MX2 Integration Marathon yes if it says "Use.." you don't have to if it says "Use.. or otherwise.."

Re: MX2 Integration Marathon sure, you can. then partial fractions from there although follow the question and use the substitution lol

Re: MX2 Integration Marathon \int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx, \; $using x = sinu$

Re: MX2 Integration Marathon lol i more or less think of trig substitutions when a question like that pops up

Re: MX2 Integration Marathon Was in my 4U test. looks very simple at first but its not lol \int secx d(tanx) = secxtanx - \int secx tan^2x dx = secxtanx - \int (sec^2x - 1) secx dx = secxtanx - \int sec^3x - secx dx \therefore 2\int sec^3x dx = secxtanx + ln|secx + tanx| \int sec^3x dx =...

\int ln(x^2 - 1)dx = xln(x^2 - 1) - \int \frac{x}{x^2 - 1} \times 2x dx = xln(x^2 - 1) - \int \frac{2x^2}{x^2 - 1}dx = xln(x^2 - 1) - \int \frac{2(x^2 - 1)}{x^2 - 1} + \frac{2}{x^2 - 1}dx Go from there EDIT: ^^^ or that. It'll probably be quicker.

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