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

InteGrand · Dec 12, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
lol a 100% magic done by substitution. anyone can find a way to play this magic:

$\int_1^3\frac{\ln x}{3+x^2}dx$

$\text{Let } I = \int_1^3\frac{\ln x}{3+x^2} \text{ d}x.$

Let u = ln x. So x = e^u ⇒ dx = e^u du.
When x = 1, u = 0, and when x = 3, u = ln 3.

$\therefore I = \int_0^{\ln 3}\frac{u\text{e}^u}{3+\text{e}^{2u}} \text{ d}u.$ .

Now, using the fact that $\int_0^{a} f(x) \text{ d}x =\int_0^{a} f(a-x) \text{ d}x$ , we have

$I = \int_0^{\ln 3}\frac{(\ln 3 - u)\text{e}^{\ln 3 - u}}{3+\text{e}^{2{(\ln 3 - u)}}} \text{ d}u.$

Since e^{ln 3 - u} = e^{ln 3}.e^-u = 3e^-u, we have

$I = \int_0^{\ln 3}\frac{(\ln 3 - u)3\text{e}^{-u}}{3+{(3\text{e}^{-u})}^2} \text{ d}u$

$= \int_0^{\ln 3}\frac{(\ln 3) \cdot 3\text{e}^{-u}}{3+9\text{e}^{-2u}} \text{ d}u - \int_0^{\ln 3}\frac{u\cdot 3\text{e}^{-u}}{3+9\text{e}^{-2u}} \text{ d}u$

$= \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u - \int_0^{\ln 3}\frac{u\text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u$

$= \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u - \int_0^{\ln 3}\frac{u\text{e}^{u}}{\text{e}^{2u}+3} \text{ d}u$ (multiplying the numerator and denominator of the integrand of the second integral on the RHS by e^2u )

$= \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u - I$

$\Rightarrow 2I = \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u.$

This is pretty easy to integrate now. Let v = e^-u ⇒ -dv = e^-u du and change the limits of integration, so

$2I = - \int_1^{\frac{1}{3}}\frac{\ln 3 }{1+3v^2} \text{ d}v$

$\Rightarrow I = \frac{1}{2}\ln 3 \int_\frac{1}{3}^1 \frac{1}{3}\cdot \frac{1}{\frac{1}{3}+v^2} \text{ d}v$

$= \frac{1}{6}\ln 3 \times \sqrt{3}\left [ \tan ^{-1}(\sqrt{3}v) \right ]_\frac{1}{3} ^1$

$= \frac{\sqrt{3}\ln 3}{6}\left \{ \tan ^{-1}\sqrt{3}-\tan ^{-1}\frac{\sqrt{3}}{3} \right \}$

$= \frac{\sqrt{3}\ln 3}{6}\left \{ \frac{\pi}{3}- \frac{\pi}{6} \right \}$

$= \frac{\pi \sqrt{3}\ln 3}{36}.$

Sy123 · Dec 12, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
Do you mean a substitution that does it all in one go?

The only thing I can think of at the moment is:

$u = \frac{x}{\sqrt{3}} \Rightarrow \ I_1$

$u = \frac{\sqrt{3}}{x} \Rightarrow \ I_2$

$2I = I_1 + I_2 = \int \dots$

Where the dots are a tan inverse integral which is easy to compute

Is that what you were looking for?

$I = \int_1^3 \frac{\ln x}{3+x^2} \ dx$

$u = \frac{x}{\sqrt{3}} , \ I_1 = \int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{\frac{1}{2} \ln 3 + \ln u}{3 + 3u^2} \cdot \sqrt{3} \ du = \frac{1}{\sqrt{3}}\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{\ln \sqrt{3} + \ln u}{1 + u^2} \ du$

$u = \frac{\sqrt{3}}{x} , \ I_2 = \frac{1}{\sqrt{3}} \int_{1 /\sqrt{3}}^{\sqrt{3}} \frac{\ln \sqrt{3} - \ln u}{1+u^2} \ du$

$2I = I_1 + I_2 = \frac{2\ln \sqrt{3}}{\sqrt{3}} \int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{du}{1+u^2} \ du$

$I = \frac{\ln 3}{4\sqrt{3}} \left(\frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi\sqrt{3}}{36} \ln 3$

InteGrand · Dec 12, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
$2I = I_1 + I_2 =\frac{\ln \sqrt{3}}{\sqrt{3}} \int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{du}{1+u^2} \ du$

Should be a $2 \ln \sqrt{3}$ in the RHS there.

FrankXie · Dec 12, 2014

Re: MX2 2015 Integration Marathon

$$ both approaches are good, i would sustitute $ x=\sqrt{3}\tan u, $ and substitute $ u=\frac{\pi}{2}-v $ to integrate $ \int_{\frac\pi6}^{\frac\pi3}\ln\tan udu$

InteGrand · Dec 23, 2014

Re: MX2 2015 Integration Marathon

$\texttt{Find}\int \frac{1}{x^7+x}\text{ d}x.$

integral95 · Dec 23, 2014

Re: MX2 2015 Integration Marathon

InteGrand said:
$\texttt{Find}\int \frac{1}{x^7+x}\text{ d}x.$

Times top and bottom by x^5

let u = x^6 and you have a standard inverse tan integral

InteGrand · Dec 23, 2014

Re: MX2 2015 Integration Marathon

integral95 said:
Times top and bottom by x^5

let u = x^6 and you have a standard inverse tan integral

Well, you actually end up with a partial fractions thing that yields a log, as the denominator becomes $u + u^2 = u(u+1)$ , numerator is 1/6.

integral95 · Dec 23, 2014

Re: MX2 2015 Integration Marathon

InteGrand said:
Well, you actually end up with a partial fractions thing that yields a log, as the denominator becomes $u + u^2 = u(u+1)$ , numerator is 1/6.

My bad

is there a faster way by any chance or was that your intended method?

InteGrand · Dec 23, 2014

Re: MX2 2015 Integration Marathon

Well, my method was to factorise the denominator as $x(x^6 +1)$ and then use the substitution you said, which gets you to the same integral as your method, so pretty much the same method.

FrankXie · Dec 23, 2014

Re: MX2 2015 Integration Marathon

NEXT QUESTION

$$ Evaluate $ \int_0^{\pi}\frac{x}{2+\sin x}dx.$

InteGrand · Dec 23, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
NEXT QUESTION

$$ Evaluate $ \int_0^{\pi}\frac{x}{2+\sin x}dx.$

$\textrm{Let } I=\int_{0}^{\pi}\frac{x}{2+\sin x}\text{ d}x$

Note that $\sin (\pi - x) = \sin x$ .

$I=\int_{0}^{\pi}\frac{\pi - x}{2+\sin (\pi-x)}\text{ d}x$

$=\int_{0}^{\pi}\frac{\pi - x}{2+\sin x}\text{ d}x$

$=\int_{0}^{\pi}\frac{\pi}{2+\sin x}\text{ d}x-\int_{0}^{\pi}\frac{x}{2+\sin x}\text{ d}x$

$=\int_{0}^{\pi}\frac{\pi}{2+\sin x}\text{ d}x-I$

$\Rightarrow 2I = \int_{0}^{\pi}\frac{\pi}{2+\sin x}\text{ d}x$

Let t = tan(x/2), where $\frac{x}{2}\in\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

Then $\sin x = \frac{2t}{1+t^2}$ and $x = 2\tan ^{-1} t$

$\text{d}x=\frac{2}{1+t^2}\text{ d}t$

t = 0 when x = 0, and $t \to \infty$ as $x\to \pi^-$ , so

$2I=\lim_{\lambda \to \infty}\int_{0}^{\lambda}\frac{\pi}{2+\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\text{ d}t$

$=\lim_{\lambda \to \infty}\int_{0}^{\lambda}\frac{2\pi}{2(1+t^2)+2t} \text{ d}t$

$=\lim_{\lambda \to \infty}\int_{0}^{\lambda}\frac{\pi}{1+t^2+t}\text{ d}t$

$=\pi\lim_{\lambda \to \infty}\int_{0}^{\lambda}\frac{1}{\frac{3}{4}+ \left ( t+\frac{1}{2} \right )^2}\text{ d}t$

$\Rightarrow 2I =\pi\times\lim_{\lambda \to \infty}\left [ \frac{2}{\sqrt{3}}\tan ^{-1}\left ( \frac{2}{\sqrt{3}}\left ( t+\frac{1}{2} \right ) \right ) \right ]_0 ^\lambda$

$\Rightarrow I=\frac{\pi}{\sqrt{3}}\times\lim_{\lambda \to \infty}\left [ \tan ^{-1}\left ( \frac{2t+1}{\sqrt{3}}\right ) \right ]_0 ^\lambda$

$=\frac{\pi}{\sqrt{3}}\left ( \frac{\pi}{2}-\tan ^{-1}\frac{1}{\sqrt{3}} \right )$

$=\frac{\pi}{\sqrt{3}}\left ( \frac{\pi}{2}-\frac{\pi}{6}\right )$

$= \frac{\pi}{\sqrt{3}}\cdot\frac{\pi}{3}$

$= \frac{\pi ^2\sqrt{3}}{9}.$

InteGrand · Dec 24, 2014

Re: MX2 2015 Integration Marathon

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

FrankXie · Dec 24, 2014

Re: MX2 2015 Integration Marathon

Improper integral again?

$$ For $ \int_0^\pi\frac{dx}{2+\sin x}, $ to avoid the singularity, split the interval into two parts and take the substitution $ u=\pi-x, \int_0^\pi\frac{dx}{2+\sin x}=\int_0^{\frac\pi2}\frac{dx}{2+\sin x}+\int_{\frac\pi2}^\pi\frac{dx}{2+\sin x}=\int_0^{\frac\pi2}\frac{dx}{2+\sin x}+\int_0^{\frac\pi2}\frac{du}{2+\sin u}=2\int_0^{\frac\pi2}\frac{dx}{2+\sin x}$

While for the last one, singularity is inevitable, you must split the interval into two pars and take respective limits.

FrankXie · Dec 24, 2014

Re: MX2 2015 Integration Marathon

As for improper integral, consider the following statement, is it true or false? Justify your answer:
$$ Because $ y=\frac{1}{x} $ is an odd function, $ \int_{-3}^3\frac{1}{x}dx=0.$

SquareZ · Dec 26, 2014

Re: MX2 2015 Integration Marathon

False because the function has an asymptote at x=0, thus the integral with boundaries which passes through the asymptote cannot be evaluated.

FrankXie · Dec 26, 2014

Re: MX2 2015 Integration Marathon

SquareZ said:
False because the function has an asymptote at x=0, thus the integral with boundaries which passes through the asymptote cannot be evaluated.

$if so, then what about $ \int_{-3}^3\frac{dx}{\sqrt[3]x}=0? $ true or false?$

InteGrand · Dec 26, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
As for improper integral, consider the following statement, is it true or false? Justify your answer:
$$ Because $ y=\frac{1}{x} $ is an odd function, $ \int_{-3}^3\frac{1}{x}dx=0.$

False because the integrals diverge when the interval is split up and appropriate limits are taken ( $\lim_{b\to 0^+}\int_{b}^{3}\frac{1}{x}\text{ d}x=\infty$ ).

FrankXie said:
$if so, then what about $ \int_{-3}^3\frac{dx}{\sqrt[3]x}=0? $ true or false?$

True, because $\lim_{b\to 0^+}\int_{b}^{3}\frac{1}{\sqrt[3]{x}}\text{ d}x$ exists (is finite), and $\lim_{b\to 0^-}\int_{-3}^{b}\frac{1}{\sqrt[3]{x}}\text{ d}x = -\lim_{b\to 0^+}\int_{b}^{3}\frac{1}{\sqrt[3]{x}}\text{ d}x$ and $\int_{-3}^{3}\frac{1}{\sqrt[3]{x}}\text{ d}x = \lim_{b\to 0^-}\int_{-3}^{b}\frac{1}{\sqrt[3]{x}}\text{ d}x + \lim_{b\to 0^+}\int_{b}^{3}\frac{1}{\sqrt[3]{x}}\text{ d}x$

Carrotsticks · Dec 27, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
Improper integral again?

$$ For $ \int_0^\pi\frac{dx}{2+\sin x}, $ to avoid the singularity, split the interval into two parts and take the substitution $ u=\pi-x, \int_0^\pi\frac{dx}{2+\sin x}=\int_0^{\frac\pi2}\frac{dx}{2+\sin x}+\int_{\frac\pi2}^\pi\frac{dx}{2+\sin x}=\int_0^{\frac\pi2}\frac{dx}{2+\sin x}+\int_0^{\frac\pi2}\frac{du}{2+\sin u}=2\int_0^{\frac\pi2}\frac{dx}{2+\sin x}$

While for the last one, singularity is inevitable, you must split the interval into two pars and take respective limits.

What singularity?

FrankXie · Dec 27, 2014

Re: MX2 2015 Integration Marathon

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

Carrotsticks · 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

But the curve 1/(2+sin x) is continuous over the interval [0,pi]?

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