HSC 2017 MX2 Integration Marathon (archive) (5 Viewers)

InteGrand · Oct 19, 2016

Re: HSC 4U Integration Marathon 2017

$\noindent Find $\int \frac{1}{x^2 -3x +2}\, \mathrm{d}x$.$

Kingom · Oct 19, 2016

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>

Kingom · Oct 19, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$\noindent Find $\int \frac{1}{x^2 -3x +2}\, \mathrm{d}x$.$

jathu123 · Oct 19, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$\noindent Find $\int \frac{1}{x^2 -3x +2}\, \mathrm{d}x$.$

$\int \frac{1}{x^2-3x+2}$ d$x = \int \frac{1}{(x-2)(x-1)} $ d$x \\ =\int( \frac{1}{x-2}-\frac{1}{x-1}) $ d$x \\\\= $ln$|x-2|-$ln$|x-1|+$C$$

calamebe · Oct 19, 2016

Re: HSC 4U Integration Marathon 2017

Paradoxica said:
$$ \noindent By very strongly considering the borders, $ \\$or otherwise, if you're masochistic, evaluate:$ \\ \int_0^1 \sin^{-1}{\sqrt{x}} \,\, \text{d}x$

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.

jathu123 · Oct 20, 2016

Re: HSC 4U Integration Marathon 2017

$$\noindent Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{1}{3+5\cos x}$ d$x$

Mahan1 · Oct 26, 2016

Re: HSC 4U Integration Marathon 2017

Kingom said:
<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

and

to simplify the inside the integrals to get

use substitution

then

let

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

Drsoccerball · Oct 26, 2016

Re: HSC 4U Integration Marathon 2017

Learn Latex pls

Drsoccerball · Oct 26, 2016

Re: HSC 4U Integration Marathon 2017

$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$

leehuan · Oct 26, 2016

Re: HSC 4U Integration Marathon 2017

Drsoccerball said:
$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$

smh...

jathu123 · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

Drsoccerball said:
$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$

$$\noindent$ I= \int \frac{x^2-1 }{(x^2+1)\sqrt{1+x^4}}$ d$x=\int \frac{x^2-1}{x(x^2+1)\sqrt{x^2+\frac{1}{x^2}}}$ d$x \\ $ dividing the denominator and the numerator by $x$ we get $\\\\ I =\int \frac{x-\frac{1}{x}}{x(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}$ d$x=\int \frac{\left ( \frac{x^2-1}{x^2} \right )}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}$ d$x \\$ let $u=x+\frac{1}{x} \Rightarrow$ d$u=1-\frac{1}{x^2}$ d$x \\ \\\therefore $ d$x=\frac{x^2}{x^2-1}$ d$u. \\\\$ Substituing in the d$x$, the numerator cancels out.$\\ I=\int \frac{1}{u\sqrt{u^2-2}}$ d$u \\\\$ let $v=\sqrt{u^2-2}\Rightarrow $ d$u = \frac{\sqrt{u^2-2}}{u}$ d$v \\$ subbing in, the integral becomes $\int \frac{1}{v^2+2}$ d$v= \frac{1}{\sqrt{2}}\tan^{-1}\frac{v}{\sqrt{2}}+C$

$$\noindent Reversing all the substitutions, we'll get \\ $I=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{u^2-2}}{\sqrt{2}}+C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}+C\\\\\\\\ $ simplifying further, it'll become $\frac{1}{\sqrt{2}}\tan^{-1}\left ( \frac{\sqrt{1+x^4}}{x\sqrt{2}} \right )+C$

abecina · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

$\int_{0}^{1}\frac{1}{(x+3)^2}\ln \left(\frac{x+2}{x+1}\right)\;\textrm{d}x$

Paradoxica · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

jathu123 said:
$$\noindent$ I= \int \frac{x^2-1 }{(x^2+1)\sqrt{1+x^4}}$ d$x=\int \frac{x^2-1}{x(x^2+1)\sqrt{x^2+\frac{1}{x^2}}}$ d$x \\ $ dividing the denominator and the numerator by $x$ we get $\\\\ I =\int \frac{x-\frac{1}{x}}{x(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}$ d$x=\int \frac{\left ( \frac{x^2-1}{x^2} \right )}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}$ d$x \\$ let $u=x+\frac{1}{x} \Rightarrow$ d$u=1-\frac{1}{x^2}$ d$x \\ \\\therefore $ d$x=\frac{x^2}{x^2-1}$ d$u. \\\\$ Substituing in the d$x$, the numerator cancels out.$\\ I=\int \frac{1}{u\sqrt{u^2-2}}$ d$u \\\\$ let $v=\sqrt{u^2-2}\Rightarrow $ d$u = \frac{\sqrt{u^2-2}}{u}$ d$v \\$ subbing in, the integral becomes $\int \frac{1}{v^2+2}$ d$v= \frac{1}{\sqrt{2}}\tan^{-1}\frac{v}{\sqrt{2}}+C$

$$\noindent Reversing all the substitutions, we'll get \\ $I=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{u^2-2}}{\sqrt{2}}+C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}+C\\\\\\\\ $ simplifying further, it'll become $\frac{1}{\sqrt{2}}\tan^{-1}\left ( \frac{\sqrt{1+x^4}}{x\sqrt{2}} \right )+C$

Somewhat convoluted.

$$\noindent$ \int \frac{x^2-1}{x^2+1} \frac{\text{d}x}{\sqrt{1+x^4}} = \int \frac{\left(1-\frac{1}{x^2}\right) \text{d}x}{\left( x+\frac{1}{x} \right) \sqrt{x^2+\frac{1}{x^2}}} \\\\ = \int \frac{\text{d} \left( x+\frac{1}{x} \right)}{\left( x+\frac{1}{x} \right) \sqrt{\left( x+\frac{1}{x} \right)^2 -2}} = \frac{1}{\sqrt{2}} \sec^{-1}{\left( \frac{x+\frac{1}{x}}{\sqrt{2}} \right)} + \mathcal{C}$

InteGrand · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

leehuan said:
smh...

Why that reaction haha? Is there some story behind this integral?

leehuan · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
Why that reaction haha? Is there some story behind this integral?

Seen it before a bit too many times

InteGrand · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

$\noindent Find $\int _{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x +\sin x}\, \mathrm{d}x$.$

InteGrand · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

$\noindent Let $\lambda , t > 0$. Show that $\int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}$, for $n=1,2,3,\ldots$.$

jathu123 · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$\noindent Find $\int _{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x +\sin x}\, \mathrm{d}x$.$

$$\noindent let $u=\frac{\pi}{2}-x \Rightarrow $ d$x = -$d$u.$ When $x=\frac{\pi}{2}, u=0.$ When $x=0, u=\frac{\pi}{2}.\\\therefore I= \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\cos x+\sin x}$ d$x=-\int_{\frac{\pi}{2}}^{0}\frac{\cos (\frac{\pi}{2}-u)}{\cos (\frac{\pi}{2}-u)+\sin (\frac{\pi}{2}-u) }$ d$u \\ = \int_{0}^{\frac{\pi}{2}}\frac{\sin u}{\sin u+\cos u}$ d$u=\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}$ d$x\\\\ \therefore 2I=\int_{0}^{\frac{\pi}{2}}\frac{ \sin x+\cos x}{\sin x + \cos x}$ d$x =\frac{\pi}{2}\\ \\ \therefore I=\frac{\pi}{4}$

Paradoxica · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$\noindent Let $\lambda , t > 0$. Show that $\int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}$, for $n=1,2,3,\ldots$.$

don't scare off the children

calamebe · Oct 28, 2016

Re: HSC 4U Integration Marathon 2017

abecina said:
$\int_{0}^{1}\frac{1}{(x+3)^2}\ln \left(\frac{x+2}{x+1}\right)\;\textrm{d}x$

http://imgur.com/a/LQn6t

Checked it on wolfram so it's all good.

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