HSC 2016 MX2 Integration Marathon (archive) (5 Viewers)

leehuan · Feb 11, 2016

Re: MX2 2016 Integration Marathon

I am ridiculously out of form

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Drsoccerball · Feb 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int_{-1}^1 \frac{x^2 e^{\tan^{-1}{x}}}{\sqrt{x^2+1}}$d$x$

Hint because I'm short on time:

$$ (1) Use IBP originally with $ u=x$

$$ (2) Integrate v by letting $ \tan{\theta} =x$

$(3) This can now be integrated by the use of two IBP's$

$(4) Now to integrate $ -u'v $ We also do integration by parts twice$

$(5) Sub in values to obtain the answer$

$Note : This is the first method that popped into my head, there's probably an easier way.$

Carrotsticks · Feb 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Hint because I'm short on time:

$$ (1) Use IBP originally with $ u=x$

$$ (2) Integrate v by letting $ \tan{\theta} =x$

$(3) This can now be integrated by the use of two IBP's$

$(4) Now to integrate $ -u'v $ We also do integration by parts twice$

$(5) Sub in values to obtain the answer$

$Note : This is the first method that popped into my head, there's probably an easier way.$

u=1/x along the the identity atan(x)+atan(1/x)=pi/2 may be fruitful for this particular integral.

Drsoccerball · Feb 12, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
u=1/x along the the identity atan(x)+atan(1/x)=pi/2 may be fruitful for this particular integral.

Yeah I didnt see this because his original integral was indefinite.

InteGrand · Feb 12, 2016

Re: MX2 2016 Integration Marathon

$\noindent Let $f(x)=\frac{\alpha}{\lambda} \left(\frac{x}{\lambda}\right)^{\alpha-1}\mathrm{e}^{-\left(\frac{x}{\lambda}\right)^{\alpha}}$ for $x>0$, where $\alpha,\lambda$ are positive constants.$

$\noindent For $x>0$, find $F(x) \equiv \int _0 ^x f(t)\text{ d}t$.$

$\noindent \textbf{Remark.} This is the cumulative distribution function (cdf) of the Weibull distribution (see link below).$

https://en.wikipedia.org/wiki/Weibull_distribution

Paradoxica · Feb 12, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Let $f(x)=\frac{\alpha}{\lambda} \left(\frac{x}{\lambda}\right)^{\alpha-1}\mathrm{e}^{-\left(\frac{x}{\lambda}\right)^{\alpha}}$ for $x>0$, where $\alpha,\lambda$ are positive constants.$

$\noindent For $x>0$, find $F(x) \equiv \int _0 ^x f(t)\text{ d}t$.$

$\noindent \textbf{Remark.} This is the cumulative distribution function (cdf) of the Weibull distribution (see link below).$

https://en.wikipedia.org/wiki/Weibull_distribution

$$\noindent By inspection, reverse chain rule, substitution or whatever you wish to use to make yourself seem pretentious, the answer is $\\F(x) \equiv 1-e^{-\left(\frac{x}{\lambda}\right)^{\alpha}}$

Paradoxica · Feb 12, 2016

Re: MX2 2016 Integration Marathon

$$a) $\int_{-1}^1 \frac{x^2 e^{\tan^{-1}{x}}}{\sqrt{x^2+1}}$d$x$

$$b) $\int_{0}^1 \frac{x^2 e^{\tan^{-1}{x}}}{\sqrt{x^2+1}}$d$x$

omegadot · Feb 13, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$a) $\int_{-1}^1 \frac{x^2 e^{\tan^{-1}{x}}}{\sqrt{x^2+1}}$d$x$

$$b) $\int_{0}^1 \frac{x^2 e^{\tan^{-1}{x}}}{\sqrt{x^2+1}}$d$x$

$$\noindent Consider the integral $I= \int^1_{-a} \frac{x^2 e^{\tan^{-1} x}}{\sqrt{x^2 + 1}} \, dx,$ where $0 \leqslant a \leqslant 1.$

$$\noindent Let $x = \tan u, dx = \sec^2 u \, du.$ Limits: $ x = 1, u = \pi/4$ and $x = -a, u = -\tan^{-1}(a) = \alpha.$ So$

$\begin{align*}I &= \int^{\frac{\pi}{4}}_\alpha \frac{\tan^2 u e^u}{\sqrt{\tan^2 u + 1}} \cdot \sec^2 \, du\\ &= \int^{\frac{\pi}{4}}_\alpha e^u \tan^2 u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_\alpha e^u (\sec^2 u - 1) \sec u \, du\\ &= \int^{\frac{\pi}{4}}_\alpha e^u \sec^3 u \, du - \int^{\frac{\pi}{4}}_\alpha e^u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_\alpha \sec^2 u \cdot e^u \sec u \, du - \int^{\frac{\pi}{4}}_\alpha e^u \sec u \, du\\ &= \left [\tan u \cdot e^u \sec u \right ]^{\frac{\pi}{4}}_\alpha - \int^{\frac{\pi}{4}}_\alpha \tan u (e^u \sec u + e^u \sec u \tan u) \, du - \int^{\frac{\pi}{4}}_\alpha e^u \sec u \, du \quad \mbox{(by parts)}\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \tan \alpha e^\alpha \sec \alpha - \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_\alpha e^u \tan^2 u \sec u \, du -\int^{\frac{\pi}{4}}_\alpha e^u \sec u \, du\end{align*}$

$$\noindent Thus$\\\begin{align*}\Rightarrow 2I &= \sqrt{2} e^{\frac{\pi}{4}} - \tan \alpha e^\alpha \sec \alpha - \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_\alpha e^u \sec u \, du\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \tan \alpha e^\alpha \sec \alpha - \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du - \left [e^u \sec u \right ]^{\frac{\pi}{4}}_\alpha + \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du \quad \mbox{(by parts)}\\&=\sqrt{2} e^{\frac{\pi}{4}} - \tan \alpha e^\alpha \sec \alpha -\sqrt{2} e^{\frac{\pi}{4}} + e^\alpha \sec \alpha\\ &= e^\alpha \sec \alpha (1 - \tan \alpha)\end{align*}$

$$\noindent Thus $I = \frac{1}{2} e^\alpha \sec \alpha (1 - \tan \alpha).$

$$\noindent For (a): $a = 1, \alpha = -\tan^{-1} (1) = -\frac{\pi}{4}$, thus $I = \sqrt{2} e^{-\frac{\pi}{4}}.$

$$\noindent For (b): $a = 0, \alpha = 0$, thus $I = \frac{1}{2}.$

omegadot · Feb 13, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^1_0 \frac{\tan^{-1} x}{x + 1}.$

Paradoxica · Feb 13, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^1_0 \frac{\tan^{-1} x}{x + 1}.$

$\noindent Integrating by parts gives $\frac{\pi}{4}\log{2}-\int_0^1 \frac{\log{(x+1)}}{x^2+1}$d$x \\ $Which is a standard integral, and that gives a final answer of $\frac{\pi}{8}\log{2} \\ $Comment: Still trying to think of a more direct method of evaluation.$

Drsoccerball · Feb 13, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent Integrating by parts gives $\frac{\pi}{4}\log{2}-\int_0^1 \frac{\log{x+1}}{x^2+1}$d$x \\ $Which is a standard integral, and that gives a final answer of $\frac{\pi}{8}\log{2} \\ $Comment: Still trying to think of a more direct method of evaluation.$

do you mean $\frac{\log{(x+1)}}{x^2+1}$ ? Also the integral according to wolfram isn't the same as yours.
EDIT: NVM they used ln not log my bad.

Paradoxica · Feb 13, 2016

Re: MX2 2016 Integration Marathon

$$\noindent By adding and subtracting $x^2$ from the numerator, evaluate $\int \frac{x^4+1}{x^6+1}$d$x$

RachelGreen · Feb 13, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
$I=\frac{1}{2^6}\int \sin^6(2x)\, dx\\ \\ = -\frac{1}{2^{12}} \int(\textrm{cis}(2x)-\textrm{cis}(-2x))^6 \, dx\\ \\=-\frac{1}{2^{12}}\int (\textrm{cis}(12x)+\textrm{cis}(-12x))-6(\textrm{cis}(8x)+\textrm{cis}(-8x))+15( \textrm{cis}(4x) +\textrm{cis}(-4x))-20 \, dx\\ \\ = -\frac{1}{2^{11}}\int \cos(12x)-6\cos(8x)+15\cos(4x)-10\, dx\\ \\ = \frac{1}{2^{11}}\left(10x-\frac{15}{4}\sin(4x)+\frac{3}{4}\sin(8x)-\frac{1}{12}\sin(12x)\right)+C.$

Can someone explain to me from step 2 and onwards? I'm a little confused? I know that cis(2x) - cis(-2x) is -2isin(2x) but like???

Drsoccerball · Feb 13, 2016

Re: MX2 2016 Integration Marathon

RachelGreen said:
Can someone explain to me from step 2 and onwards? I'm a little confused? I know that cis(2x) - cis(-2x) is -2isin(2x) but like???

$Instead of using demoivre's theorem straight away that is : $ (rcis{\theta})^n= r^ncis{(n\theta)} $ He used binomial expansion and then used demoivre's theorem.$

$$ You're allowed to do this because an even power removes the imaginary term and thus allows the integral to be integrated. ( Obviously in complex integrals we can separate the imaginary terms and real and integrate separately but that's not HSC level.)$

Drsoccerball · Feb 14, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent By adding and subtracting $x^2$ from the numerator, evaluate $\int \frac{x^4+1}{x^6+1}$d$x$

$\int{\frac{x^4-x^2+1}{x^6+1} +\frac{x^2}{x^6+1}}dx$

$\tan ^{-1}x + \frac{1}{3} \tan ^{-1} (x^3) +C$

Paradoxica · Feb 14, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{\log(1-e^x)}{e^{2x}}$d$x$

Drsoccerball · Feb 14, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\log(1-e^x)}{e^{2x}}$d$x$

The straight forward way is to just let $u=\log{1-e^x}$ and the use IBP. But im guessing there's an easier way? Like reverse quotient rule +- a constant ?

Paradoxica · Feb 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
The straight forward way is to just let $u=\log{1-e^x}$ and the use IBP. But im guessing there's an easier way? Like reverse quotient rule +- a constant ?

That substitution looks unwieldy. Are you sure that's easier than doing a normal substitution?

The fastest way my mind suggests is to integrate by parts directly to remove the logarithm, then perform the obvious substitution OR simplify then perform the less obvious substitution.

Drsoccerball · Feb 14, 2016

Re: MX2 2016 Integration Marathon

New Question:

$\int_0^{\infty}{\frac{x \ln{x}}{(a^2+x^2)(a^2x^2+1)}}dx$

RachelGreen · Feb 14, 2016

Re: MX2 2016 Integration Marathon

I've just started integration 4u, and am looking through this read from the beginning. How do you guys do Integration by Parts so quickly? I'm sick and tired of writing u and u' and v' and v

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