HSC 2017 MX2 Integration Marathon (archive) (1 Viewer)

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Kingom

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Re: HSC 4U Integration Marathon 2017

<a href="https://www.codecogs.com/eqnedit.php?latex=\int&space;\frac{sinx&space;-&space;cosx}{(sinx&plus;cosx)\sqrt{sinxcosx&space;&plus;&space;sin^2xcos^2x}}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\int&space;\frac{sinx&space;-&space;cosx}{(sinx&plus;cosx)\sqrt{sinxcosx&space;&plus;&space;sin^2xcos^2x}}" title="\int \frac{sinx - cosx}{(sinx+cosx)\sqrt{sinxcosx + sin^2xcos^2x}}" /></a>
 

Kingom

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Re: HSC 4U Integration Marathon 2017

<a href="https://www.codecogs.com/eqnedit.php?latex=\inline&space;\int&space;\frac{1}{x^2-3x&plus;2}&space;=\int&space;\frac{1}{(x-1)(x-2)}&space;=&space;\int&space;\frac{1}{x-2}&space;-&space;\int&space;\frac{1}{x-1}&space;=&space;\ln&space;(x-2)&space;-&space;\ln&space;(x-1)&space;&plus;C" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\inline&space;\int&space;\frac{1}{x^2-3x&plus;2}&space;=\int&space;\frac{1}{(x-1)(x-2)}&space;=&space;\int&space;\frac{1}{x-2}&space;-&space;\int&space;\frac{1}{x-1}&space;=&space;\ln&space;(x-2)&space;-&space;\ln&space;(x-1)&space;&plus;C" title="\int \frac{1}{x^2-3x+2} =\int \frac{1}{(x-1)(x-2)} = \int \frac{1}{x-2} - \int \frac{1}{x-1} = \ln (x-2) - \ln (x-1) +C" /></a>
 

calamebe

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Re: HSC 4U Integration Marathon 2017

Also, a way to do this is note that arcsin(root(0))=0, and so this can be evaluated using the integral from 0 to a of a function of x = ab - the integral from 0 to b of the inverse function of x, where f(a)=b. I had to type this like I just did sorry haha.
 

Mahan1

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Re: HSC 4U Integration Marathon 2017

<a href="https://www.codecogs.com/eqnedit.php?latex=\int&space;\frac{sinx&space;-&space;cosx}{(sinx+cosx)\sqrt{sinxcosx&space;+&space;sin^2xcos^ 2x}}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\int&space;\frac{sinx&space;-&space;cosx}{(sinx+cosx)\sqrt{sinxcosx&space;+&space;sin^2xcos^ 2x}}" title="\int \frac{sinx - cosx}{(sinx+cosx)\sqrt{sinxcosx + sin^2xcos^2x}}" /></a>
you can rearrange gif.latex.gifand gif.latex.gif to simplify the inside the integrals to get

gif.latex.gif
use substitution gif.latex.gif
then gif.latex.gif

gif.latex.gif

gif.latex.gif
let gif.latex.gif

gif.latex.gif constant

From the you can see the integrals is tan inverse of \sqrt{u^2-1}
We can back track to find it in terms of theta
 

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