HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

seanieg89 · Dec 27, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
the integrand has a discontinuity (either a finite jump or infinity) in the interval of integration or at either endpoint of the interval, like tangent of x in any interval containing \pi/2, 1/x in any interval containing 0

Where is $\frac{1}{2+\sin(x)}$ discontinuous?

FrankXie · Dec 27, 2014

Re: MX2 2015 Integration Marathon

oh my bad, I really meant the solution given by InteGrand, where the substitution introduced a singularity. my solution does not contain any singularity

juantheron · Jan 7, 2015

Re: MX2 2015 Integration Marathon

$Evaluation\; of\; $\displaystyle \int \frac{4x^5-1}{(x^5+x+1)^2}dx$\; and $\displaystyle \int \sqrt{1+\tan x \cdot \tan (x+\alpha)}dx$

juantheron · Jan 7, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$\mathfrak{Evaluate}\int \left ( \frac{1}{x^7+x} \right )\text{ d}x$

$Let $\displaystyle \mathcal{I} = \frac{1}{x^7+x}dx = \int\frac{x^{-7}}{1+x^{-6}}dx$

$Now \; let $(1+x^{-6})dx = t\;,$ Then \; $\displaystyle x^{-7}dx = -\frac{1}{6}dt$

$So\; Integral $\displaystyle I = -\frac{1}{6}\int \frac{1}{t}dt = -\frac{1}{6}\ln \left|t\right|+\mathcal{C}$

$So \; $\displaystyle \mathcal{I} = \frac{1}{x^7+x}dx = \frac{1}{6}\ln \left|\frac{x^6}{1+x^6}\right|+\mathcal{C}$

juantheron · Jan 7, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$\mathfrak{Evaluate}\int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x$

$$Let \; $\displaystyle \mathcal{I} = \int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x$ = \displaystyle \left[\frac{1}{1+2^{\frac{1}{x}}}\right]_{-1}^{1} = \displaystyle \left[\left(\frac{1}{1+2^{-1}}\right)-\left(\frac{1}{1+2}\right)\right]$

$So \; $\displaystyle \mathcal{I} = \int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x = \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$

FrankXie · Jan 7, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$$Let \; $\displaystyle \mathcal{I} = \int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x$ = \displaystyle \left[\frac{1}{1+2^{\frac{1}{x}}}\right]_{-1}^{1} = \displaystyle \left[\left(\frac{1}{1+2^{-1}}\right)-\left(\frac{1}{1+2}\right)\right]$

$So \; $\displaystyle \mathcal{I} = \int_{-1}^{1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{1+2^{\frac{1}{x}}} \right )\text{ d}x = \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$

This is incorrect, because you did not notice the jump at x=0

FrankXie · Jan 8, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$Evaluation\; of\; $\displaystyle \int \frac{4x^5-1}{(x^5+x+1)^2}dx$\; and $\displaystyle \int \sqrt{1+\tan x \cdot \tan (x+\alpha)}dx$

For the first integral I, reverse the quotient rule:
$\frac{4x^5-1}{(x^5+x+1)^2}=\frac{(-x^5-x-1)+x(5x^4+1)}{(x^5+x+1)^2}=\frac{(x^5+x+1)(-x)^\prime-(-x)(x^5+x+1)^\prime}{(x^5+x+1)^2}=\left(\frac{-x}{x^5+x+1}\right)^\prime. $ Therefore, $ I=-\frac{x}{x^5+x+1}+c.$

porcupinetree · Jan 16, 2015

Re: MX2 2015 Integration Marathon

Find $\int\sqrt{\tan x} \, dx$

juantheron · Jan 24, 2015

Re: MX2 2015 Integration Marathon

$Let\; $I = \sqrt{\tan x}\;\mathrm{d}x$\; and \; $J = \sqrt{\cot x}\;\mathrm{d}x$ .

$\displaystyle I + J = \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x$

$\displaystyle = \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x$

$\displaystyle = \sqrt{2} \int\frac{(\sin x - \cos x)'}{\sqrt{1-(\sin x - \cos x)^2}} \;\mathrm{d}x$

$\displaystyle = \sqrt{2} \sin^{-1}(\sin x - \cos x) + \mathcal{C^{'}}$

$and \; $I - J = \int\left(\sqrt{\tan x} - \sqrt{\cot x}\right) \;\mathrm{d}x$

$\displaystyle = \sqrt{2} \int\frac{(\sin x - \cos x)}{\sqrt{\sin 2x}} \;\mathrm{d}x$

$\displaystyle = -\sqrt{2} \int\frac{(\sin x + \cos x)'}{\sqrt{(\sin x + \cos x)^2 - 1}} \;\mathrm{d}x$

$\displaystyle = -\sqrt{2} \ln\left|(\sin x + \cos x) + \sqrt{(\sin x + \cos x)^2 - 1}\right| + \mathcal{C^{''}}$

$$ Now, \; adding \; $(1)$\; and\; $(2)$:$

$So $I\displaystyle = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathcal{C}$

juantheron · Jan 24, 2015

Re: MX2 2015 Integration Marathon

$Evaluation of $\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$

Sy123 · Jan 25, 2015

Re: MX2 2015 Integration Marathon

$\\ \int \frac{\sin (3x)}{\sin x} \ dx$

RealiseNothing · Jan 25, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \int \frac{\sin (3x)}{\sin x} \ dx$

We can cancel out the $x$ and $\sin$ so that we are left with:

$\int 3~dx$

$=3x+C$

InteGrand · Jan 27, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \int \frac{\sin (3x)}{\sin x} \ dx$

$\frac{\sin 3x}{\sin x}=\frac{\sin (2x+x)}{\sin x}$

$=\frac{\sin 2x \cos x + \cos 2x \sin x}{\sin x}$

$=\frac{2 \sin x\cos^2 x }{\sin x}+\cos 2x$

$=2\cos^2 x + 1 - 2\sin^2 x$

$=1+2(\cos^2 x - \sin^2 x)$

$=1+2\cos 2x$ .

$\therefore \int \frac{\sin 3x}{\sin x} $ d$x=\int (1+2\cos 2x) $ d$x$

$= x+\sin 2x + C$ .

InteGrand · Jan 27, 2015

Re: MX2 2015 Integration Marathon

$\mathfrak{Find} \int \frac{\ln(x+1)}{x^2} $ d$x.$

leehuan · Feb 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$\mathfrak{Find} \int \frac{\ln(x+1)}{x^2} $ d$x.$

$This\quad integral\quad has\quad a\quad domain\quad restriction:\quad x>-1\\ \int { \frac { \ln { \left( x+1 \right) } }{ { x }^{ 2 } } dx } \\ IBP:\quad u=\ln { \left( x+1 \right) } ,\quad u\prime =\frac { 1 }{ x+1 } ,\quad v\prime =\frac { 1 }{ { x }^{ 2 } } ,\quad v=-\frac { 1 }{ x } \\ =-\frac { \ln { \left( x+1 \right) } }{ x } +\int { \frac { 1 }{ x\left( x+1 \right) } dx } \\ Now,\quad let\quad \frac { 1 }{ x\left( x+1 \right) } \equiv \frac { A }{ x } +\frac { B }{ x+1 } \\ 1\equiv A(x+1)+Bx\\ When\quad x=0,\quad we\quad have\quad 1=A\\ When\quad x=-1\quad we\quad have\quad -1=B$
$\ \therefore \frac { 1 }{ x\left( x+1 \right) } \equiv \frac { 1 }{ x } -\frac { 1 }{ x+1 } \\ Hence\quad the\quad integral\quad is\quad now\\ -\frac { \ln { \left( x+1 \right) } }{ x } +\int { \left( \frac { 1 }{ x } -\frac { 1 }{ x+1 } \right) dx } \\ =-\frac { \ln { \left( x+1 \right) } }{ x } +\ln { \left| x \right| } -\ln { \left( x+1 \right) } +C$

InteGrand · Feb 6, 2015

Re: MX2 2015 Integration Marathon

New question

$\text{Find } I\equiv \int \sqrt{1+\text{e}^x}\text{ d}x.$

HeroicPandas · Feb 6, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
New question

$\text{Find } I\equiv \int \sqrt{1+\text{e}^x}\text{ d}x.$

Let $u^2 = 1+e^x$ and a little side-note that $1+e^x > 0$ so $|u| > 0$

So, $2udu = e^x dx$

$I = 2 \int \frac{u^2}{u^2 - 1} du$

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

$I = 2 \int ( 1 - \frac{0.5}{u+1} + \frac{0.5}{u-1} )du$

$I = 2 \sqrt{1+e^x} - ln(1+\sqrt{1+e^x}) + ln( \sqrt{1+e^x} +1) +C$

Question:

$$Find $I = \int \frac{sec^2 (x)}{9 + tan^2 (x)} dx$

hypermax · Feb 7, 2015

Re: MX2 2015 Integration Marathon

HeroicPandas said:
Let $u^2 = 1+e^x$ and a little side-note that $1+e^x > 0$ so $|u| > 0$

So, $2udu = e^x dx$

$I = 2 \int \frac{u^2}{u^2 - 1} du$

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

$I = 2 \int ( 1 - \frac{0.5}{u+1} + \frac{0.5}{u-1} )du$

$I = 2 \sqrt{1+e^x} - ln(1+\sqrt{1+e^x}) + ln( \sqrt{1+e^x} +1) +C$

Question:

$$Find $I = \int \frac{sec^2 (x)}{9 + tan^2 (x)} dx$

Is that necessary when your doing an exam?

braintic · Feb 7, 2015

Re: MX2 2015 Integration Marathon

HeroicPandas said:
Question:

$$Find $I = \int \frac{sec^2 (x)}{9 + tan^2 (x)} dx$

A little easy compared to the standard set here

integral95 · Feb 7, 2015

Re: MX2 2015 Integration Marathon

hypermax said:
Is that necessary when your doing an exam?

No the side note he wrote won't make a difference

HeroicPandas said:
$$Find $I = \int \frac{sec^2 (x)}{9 + tan^2 (x)} dx$

By inspection

$I = \frac{1}{3}tan^{-1}(\frac{tanx}{3})+ c$

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HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

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