HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

juantheron · Apr 19, 2014

Re: MX2 Integration Marathon

$\displaystyle \int\frac{1}{(x+\alpha)^{1+\frac{1}{n}}\cdot (x+\beta)^{1-\frac{1}{n}}}dx = \int \frac{1}{\left(\frac{x+\alpha}{x+\beta}\right)^{1+\frac{1}{n}}}\cdot \frac{1}{(x+\beta)^2}dx$

$Now \; let \; $\displaystyle \left(\frac{x+\alpha}{x+\beta}\right) = t$\;,then \; $\displaystyle \frac{1}{(x+\beta)^2}dx = \frac{1}{(\beta-\alpha)}dt$

$So\ Integral \; convert \; into \; $\displaystyle \frac{1}{(\beta-\alpha)}\int \frac{1}{t^{1+\frac{1}{n}}}dt = \frac{1}{(\beta-\alpha)}\int t^{-1-\frac{1}{n}}dt$

$\displaystyle = \frac{1}{(\beta-\alpha)}\cdot \frac{t^{-\frac{1}{n}}}{-\frac{1}{n}}+\mathbb{C} = \frac{1}{(\beta-\alpha)}\cdot -n \cdot \frac{1}{t^{\frac{1}{n}}}+\mathbb{C}$

$where \; $\displaystyle t = \left(\frac{x+\alpha}{x+\beta}\right)$

juantheron · Apr 19, 2014

Re: MX2 Integration Marathon

$\displaystyle I = \int \frac{1}{\sqrt{x}\cdot \left(\sqrt[4]{x}+1\right)^{10}}dx$

$\displaystyle J = \int\frac{1+x^2}{\left(1-x^2\right)\sqrt{1+x^4}}dx$

$\displaystyle K = \int\frac{1}{\left(1-x^2\right)\sqrt{1+x^4}}dx$

$\displaystyle L = \int \frac{\cos^2 x}{1+\cos^2 x}dx$

$\displaystyle M = \int \frac{1+\tan^2 x}{\left(4+\tan^2 x\right)\cdot \tan^3 x}dx$

Sy123 · Apr 19, 2014

Re: MX2 Integration Marathon

juantheron said:
$\displaystyle I = \int \frac{1}{\sqrt{x}\cdot \left(\sqrt[4]{x}+1\right)^{10}}dx$

$\displaystyle J = \int\frac{1+x^2}{\left(1-x^2\right)\sqrt{1+x^4}}dx$

$\displaystyle K = \int\frac{1}{\left(1-x^2\right)\sqrt{1+x^4}}dx$

$\displaystyle L = \int \frac{\cos^2 x}{1+\cos^2 x}dx$

$\displaystyle M = \int \frac{1+\tan^2 x}{\left(4+\tan^2 x\right)\cdot \tan^3 x}dx$

$\\x = (u-1)^4 \\ \Rightarrow \ I = \int \frac{4(u-1)^3}{(u-1)^2 u^{10}} \ du = 4\int u^{-9} - u^{-10} \ du = \frac{4}{9u^9} - \frac{1}{2u^{8}} + c \ \ \rightarrow \ u = \sqrt[4]{x} + 1$

---------

$\\ u = x - \frac{1}{x} \\ \Rightarrow \ J = \int \frac{du}{-u \sqrt{u^2-2}} = - \int \frac{u}{u^2 \sqrt{u^2-2}} \ du \\ m^2 = u^2-2 \\ \Rightarrow \ J = -\int \frac{dm}{m^2+2} = -\frac{1}{\sqrt{2}} \tan^{-1} \frac{m}{\sqrt{2}} + c \ \ \rightarrow \ m = \sqrt{x^2 + \frac{1}{x^2}}$

---------

$\\ $let$ \ \alpha_K = \int \frac{1-x^2}{(1-x^2)\sqrt{1+x^4}} \ dx \ \ \Rightarrow \ \alpha_K + J = 2K \\ \alpha_K = \int \frac{1}{\sqrt{1+x^4}} \ dx \ $which cannot be found in elementary form$$

--------

$\\ L = \int 1 - \frac{1}{1+ \cos^2x} \ dx = x - \int \frac{\sec^2x}{2 + \tan^2{x}} \ dx = x - \frac{1}{\sqrt{2}} \tan^{-1} \frac{\tan x}{\sqrt{2}} + c$

-------

$\\ u = \tan x \\ \Rightarrow \ M = \int \frac{du}{u^3(4+u^2)} = \int \frac{1}{4u^3} - \frac{1}{16u} + \frac{1}{32} \frac{2u}{4+u^2} \ du = \frac{1}{32}\ln (4 + \tan^2x) - \frac{1}{16} \ln \tan x - \frac{1}{8\tan^2{x}} + c$

Sy123 · Apr 19, 2014

Re: MX2 Integration Marathon

$\int \tan (x) \cdot \tan (2x) \cdot \tan (3x) \ dx$

dunjaaa · Apr 19, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-19 at 9.07.36 PM.png

-> Question

Davo_01 · Apr 19, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int \tan (x) \cdot \tan (2x) \cdot \tan (3x) \ dx$

$\tan (x) \cdot \tan (2x) \cdot \tan (3x)$

$=\tan (x) \cdot \tan (2x) \cdot (\dfrac{\tan(x)+\tan(2x)}{1-\tan (x) \cdot \tan (2x) })$

$=-(1-\tan (x) \cdot \tan (2x) \cdot) (\dfrac{\tan(x)+\tan(2x)}{1-\tan (x) \cdot \tan (2x) }+\tan(3x)$

$=\tan(3x)-(\tan(x)+\tan(2x))$

Which is easy to integrate...
nice question btw

Davo_01 · Apr 19, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30312 -> Question

$\dfrac{(x+1)^2 (x^2-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}}$

Dividing by $x^3$ (Note this is assuming $x>0)$

$\dfrac{(\dfrac{x^2+2x+1}{x}) (1-\dfrac{1}{x^2})}{(x+\dfrac{1}{x})^2 \sqrt{x^2+1+\dfrac{1}{x^2}}}$

$\therefore \int \dfrac{(x+1)^2 (x^2-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} dx$

$= \int \dfrac{(x+2+\dfrac{1}{x}) (1-\dfrac{1}{x^2})}{(x+\dfrac{1}{x})^2 \sqrt{x^2+1+\dfrac{1}{x^2}}} dx$

Use substitution $\sec(z)=x+\dfrac{1}{x}$

$\sec(z) \tan(z) \cdot dz =(1-\dfrac{1}{x^2})\cdot dx$

$= \int \dfrac{(\sec(z)+2) (\sec(z) \tan(z)) \cdot dz}{(\sec(z))^2 \sqrt{\sec^2 (z)-1}}$

$= \int \dfrac{\sec(z)+2}{\sec(z)} \cdot dz$

$=z+2\sin(z)+c$

Where $\sec(z)=x+\dfrac{1}{x}$ and substitute back into x...

dunjaaa · Apr 20, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-20 at 12.05.50 AM.png

Made this one up

Davo_01 · Apr 20, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30314 Made this one up

Using the substitution $u=\frac{ln(x)}{x}$ yields the integral:

$\int \dfrac{du}{u^4+1}$

$=\dfrac{\sqrt{2}}{8}\int \dfrac{2u+2\sqrt{2}}{u^2+u\sqrt{2}+1}-\dfrac{2u-2\sqrt{2}}{u^2-u\sqrt{2}+1} \cdot du$ (Partial Fractions)

$=\dfrac{\sqrt{2}}{8}ln(\dfrac{u^2+u\sqrt{2}+1}{u^2-u\sqrt{2}+1})$
$+\dfrac{1}{4}\int\dfrac{1}{(u+\frac{1}{\sqrt{2}})^2+\frac{1}{2}}+\dfrac{1}{(u-\frac{1}{\sqrt{2}})^2+\frac{1}{2}} \cdot du$

$=\dfrac{\sqrt{2}}{8}ln(\dfrac{u^2+u\sqrt{2}+1}{u^2-u\sqrt{2}+1})$
$+\dfrac{\sqrt{2}}{4}[\tan^{-1}(u\sqrt{2}+1)+\tan^{-1}(u\sqrt{2}-1)]+c$

$=\dfrac{\sqrt{2}}{8}(ln(\dfrac{u^2+u\sqrt{2}+1}{u^2-u\sqrt{2}+1})$ $+2 \tan^{-1}(\dfrac{u\sqrt{2}}{1-u^2}))+c$

Where $u=\frac{ln(x)}{x}$ and substitute back in terms of x...

dunjaaa · Apr 20, 2014

Re: MX2 Integration Marathon

pretty much what i did

, after simplifying you should get something like

Screen shot 2014-04-20 at 1.35.41 AM.png

Davo_01 · Apr 20, 2014

Re: MX2 Integration Marathon

dunjaaa said:
pretty much what i did , after simplifying you should get something like View attachment 30315

Yeh sweet

juantheron · Apr 21, 2014

Re: MX2 Integration Marathon

$\displaystyle \int\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx$

$\displaystyle \int x^{26}\cdot \left(x-1\right)^{17}\cdot (5x-3)dx$

juantheron · Apr 21, 2014

Re: MX2 Integration Marathon

$I\; have \ tried \; like \; this \; way....$

$\displaystyle \int \tan x\cdot \tan 2x\cdot \tan 3xdx = \int \tan x \cdot \frac{2\tan x}{1-\tan^2 x}\cdot \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}dx$

$Now\; Let \; $\tan x= t\;,$ Then \; $\displaystyle \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+t^2}dt$

$\displaystyle \int t\cdot \frac{2t}{1-t^2}\cdot \frac{3t-t^3}{1-3t^2}\cdot \frac{1}{1+t^2}dt = \int \frac{2t^3\cdot (3-t^2) }{(1-t^2)\cdot (1+t^2)\cdot (1-3t^2)}dt$

$Now \; let\; $t^2=u\;\;,$ then \; $2tdt = du$

$\displaystyle \int\frac{u\cdot (3-u)}{(1-u)\cdot(1+u)\cdot(1-3u)}du = \frac{1}{3}\int\frac{3u-u^2}{(u-1)\cdot (u+1)\cdot(u-\frac{1}{3})}du$

$Now\; using\; partial \; fraction....$

Davo_01 · Apr 21, 2014

Re: MX2 Integration Marathon

juantheron said:
$\displaystyle \int\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx$

$\displaystyle \int x^{26}\cdot \left(x-1\right)^{17}\cdot (5x-3)dx$

Nice question, works out very neatly

$\int\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx$

Let $t=\tan{x}$

$\therefore \dfrac{dt}{1+t^2}=dx$

$=\int\frac{(\dfrac{2t}{\sqrt{1+t^2}})^2 (\dfrac{1}{\sqrt{1+t^2}})2 }{((\dfrac{2t}{\sqrt{1+t^2}})^3 +(\dfrac{1}{\sqrt{1+t^2}})^3 )^2}\cdot \dfrac{dt}{1+t^2}$

$=\int \dfrac{4t^2}{(8t^3+1)^2} \cdot dt$

$=-\dfrac{1}{6(8\tan^3{x}+1)} +c$

For 2nd question so far i can only think of expanding ahahaha, which isnt to long if we leave it in sigma notation

dunjaaa · Apr 21, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-21 at 6.33.58 PM.png

-> question

Sy123 · Apr 21, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30321 -> question

$\\ u = -x \\ \Rightarrow \ \int_{-\pi/4}^{\pi/4} \frac{e^x}{1+e^x}\sec^3x \ dx = \int_{-\pi/4}^{\pi/4} \frac{1}{1+e^u}\sec^3 u \ du \\ \Rightarrow \ 2I = \int_{-\pi/4}^{\pi/4} \sec^3 x \ dx \Rightarrow \ \dots \dots$

dunjaaa · Apr 21, 2014

Re: MX2 Integration Marathon

Yeah pretty much, was aiming to utilise the property int{(x=-a to x=a) f(x)dx} = int{(x=0 to x=a) f(x)+f(-x)dx}

dunjaaa · Apr 30, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-30 at 8.42.52 PM.png

Sy123 · May 4, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30372

$\\ x = \sin \theta \\ \Rightarrow \ I = \int \ln(1 + \cos \theta) \cos \theta \ d\theta \\ \\ I = \sin \theta \ln(1+\cos \theta) + \underbrace{\int \frac{\sin^2 \theta}{1+\cos \theta} d\theta}_{J} \\ \\ J = \int \frac{1-\cos^2 \theta}{1+ \cos \theta} d\theta = \int 1 - \cos \theta = \theta - \sin \theta \\ \\ \therefore \ I = \sin^{-1}(x) - x + x\ln(1+\sqrt{1-x^2}) + c$

Sy123 · May 4, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi/2} \frac{\sin^2(n\theta)}{\sin^2 \theta} \ d\theta$

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