HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

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

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

HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

Active Member

Active Member

This too shall pass

This too shall pass

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Attachments

Active Member

Active Member

Active Member

Active Member

Active Member

This too shall pass

Active Member

Active Member

This too shall pass

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 4)