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

kawaiipotato · Oct 12, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$Integrate without substitution.$
$$a)$\int {\frac{2x^3-1}{x^4+x}}dx$
$$b)$\int {\frac{dx}{x^2\sqrt{x^2+4}}}$

b)
$$ I $ = \int \frac{x^{2}}{x^4 + 4x^{2}} dx + \int \frac{4}{x^4 + 4x^{2}} dx$
= $\int \frac{1}{x^2 + 4} dx + \int \frac{4}{(x^2 + 2)^2 - 4} dx$
= $\frac{1}{2}tan^{-1} \left ( \frac{x}{2} \right ) - \ln \left ( \frac{x^2 +4}{x^{2}} \right ) + \mathcal{C}$
Nvm it forgot it had a square root at the top

leehuan · Oct 13, 2015

Re: MX2 2015 Integration Marathon

a) is just partial fractions.

$\int { \frac { 2{ x }^{ 3 }-1 }{ x\left( { x }^{ 3 }+1 \right) } dx } \\ =\int { \left( \frac { 2x-1 }{ { x }^{ 2 }-x+1 } -\frac { 1 }{ x } +\frac { 1 }{ x+1 } \right) dx } \\ =\ln { \left( { x }^{ 2 }-x+1 \right) } -\ln { \left| x \right| } +\ln { \left| x+1 \right| } +C$

But if the reverse chain rule constitutes as integration by substitution then I cannot see any means in solving the question except for an unreliable integration by parts.

Paradoxica · Oct 13, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
a) is just partial fractions.

$\int { \frac { 2{ x }^{ 3 }-1 }{ x\left( { x }^{ 3 }+1 \right) } dx } \\ =\int { \left( \frac { 2x-1 }{ { x }^{ 2 }-x+1 } -\frac { 1 }{ x } +\frac { 1 }{ x+1 } \right) dx } \\ =\ln { \left( { x }^{ 2 }-x+1 \right) } -\ln { \left| x \right| } +\ln { \left| x+1 \right| } +C$

But if the reverse chain rule constitutes as integration by substitution then I cannot see any means in solving the question except for an unreliable integration by parts.

Reverse Chain Rule is not the same thing as substitution. From the moment you arbitrarily substitute, you have given up the more elegant path. However, I can think of a case where substitution would be more elegant, rather than less.

Paradoxica · Oct 16, 2015

Re: MX2 2015 Integration Marathon

$\int \frac{dx}{x(1+x^3)(1+3x^3)}$

porcupinetree · Oct 16, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int \frac{dx}{x(1+x^3)(1+3x^3)}$

Nice Q.

turnerloos · Oct 18, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Nice Q.

could you explain the "there is no need..." part? I don't understand your reason behind it

Paradoxica · Oct 18, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Nice Q.

I think substitution would have been faster this time around.
o0o

omegadot · Oct 18, 2015

Re: MX2 2015 Integration Marathon

$\int^1_0 \frac{\ln (1 + x)}{1 + x^2} \mbox{d}x$

Drsoccerball · Oct 18, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$\int^1_0 \frac{\ln (1 + x)}{1 + x^2} \mbox{d}x$

Im pretty sure this integral has been solved at least 3 times.

omegadot · Oct 18, 2015

Re: MX2 2015 Integration Marathon

$$Okay. Sorry about that. How about$\\ \int^3_{\frac{1}{3}} \frac{\tan^{-1} x}{x^2 - x + 1} \mbox{d}x$

InteGrand · Oct 18, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$Okay. Sorry about that. How about$\\ \int^3_{\frac{1}{3}} \frac{\tan^{-1} x}{x^2 - x + 1} \mbox{d}x$

$Just from the bounds, I would guess a substitution of $u=\frac{1}{x}$.$

Drsoccerball · Oct 18, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Just from the bounds, I would guess a substitution of $u=\frac{1}{x}$.$

Yes you end up with :

$I= \frac{\pi}{4} \int_{\frac{1}{3}}^3{\frac{du}{u^2-u+1}}$

$Which can be integrated via completion of the square etc...$

omegadot · Oct 18, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Yes you end up with :

$I= \frac{\pi}{4} \int_{\frac{1}{3}}^3{\frac{du}{u^2-u+1}}$

$Which can be integrated via completion of the square etc...$

Correct! How about:
$\int \left (\frac{x}{x\sin x + \cos x} \right )^2\mbox{d}x$

juantheron · Oct 22, 2015

Re: MX2 2015 Integration Marathon

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

$We can write $\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\}$

$= \sqrt{1+x^2}\cdot \cos \left(x-\phi \right)$

$So Here $\displaystyle \cos \phi = \frac{1}{\sqrt{1+x^2}}$ and $\displaystyle \sin \phi = \frac{x}{\sqrt{1+x^2}}$

$and $\tan \phi = x\Rightarrow \phi = \tan^{-1}(x)$

$So Integral $\displaystyle I = \int\frac{x^2}{(1+x^2)\cdot \cos^2(x-\phi)}dx$

$\int \sec^2 (x-\phi)\cdot \frac{x^2}{1+x^2} dx = \int \sec^2 (x-\tan^{-1}(x))\cdot \frac{x^2}{(1+x^2)}dx$

$Now Let $\left(x-\tan^{-1}x\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{1+x^2}\right)dx=dt$

$\frac{x^2}{1+x^2}dx = dt$$

$So $\displaystyle I = \int \sec^2 t dt = \tan t +\mathcal {C} = \tan \left(x-\tan^{-1} x\right)+\mathcal{C} = \left(\frac{\tan x-x}{1+x\cdot \tan x}\right)+\mathcal{C}$

$So $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx = \left(\frac{\sin x- x\cos x}{\cos x+x\sin x}\right)+\mathcal{C}$

Paradoxica · Oct 22, 2015

Re: MX2 2015 Integration Marathon

$\int {\tan^2{(x-a)}\tan^2{(x-b)}}dx$

juantheron · Oct 22, 2015

Re: MX2 2015 Integration Marathon

$Let $I = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx\;,$ Now put $x=\frac{1-t}{1+t} = -\left(\frac{t-1}{t+1}\right) = \frac{2}{1+t}-1$

$So we get $dx = -\frac{2}{(1+t)^2}dt\;,$ and changing limit\;, we get$

$I = \int_{1}^{0}\frac{\ln\left(\frac{2}{1+t}\right)}{2(1+t^2)}\cdot (1+t)^2\cdot \frac{-2}{(1+t)^2}dt = \int_{0}^{1}\left[\frac{\ln (2)-\ln(1+t)}{1+t^2}\right]dt$

$So we get $I = \ln (2)\int_{0}^{1}\frac{1}{1+t^2}dt - \int_{0}^{1}\frac{\ln (1+t)}{1+t^2}dt = \ln (2)\cdot \frac{\pi}{4}-\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx$

$So $I = \frac{\pi\cdot \ln(2)}{4}-I \Rightarrow 2I = \frac{\pi\cdot \ln(2)}{4}\Rightarrow I = \frac{\pi\cdot \ln(2)}{8}$

juantheron · Oct 22, 2015

Re: MX2 2015 Integration Marathon

$(1)\; If $f(x) = x^3+3x+4\;,$ Then value of $\int_{-1}^{1}f(x)dx+\int_{0}^{4}f^{-1}(x)dx =$

$(2)\; Let $I_{n} = \int_{0}^{\frac{\pi}{2}}x(\cos x+\sin x)^ndx\;,$ Then value of $\frac{101I_{101}-\frac{\pi}{2}}{I_{99}} =$

$(3)\; Let $f(k) = \int_{-1}^{1}\frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}dx.$ Then $\sum_{k=0}^{99}f(k) =$

juantheron · Oct 22, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
In (c) your integral does not depend on 'r'. Should that be V_n?

Sorry Manager I have edited my post.

InteGrand · Oct 22, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
Sorry Manager I have edited my post.

Here's the question (part (c)) if anyone wants to see it and doesn't want to go back and find it (parts (a) and (b) already done):

juantheron said:
$(a)\;\; \int_{0}^{1}\binom{207}{7}x^{200}\cdot (1-x)^{7}dx\;\;\;\;\; (b)\;\; \int_{0}^{\ln 2}\frac{2e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^x+1}dx$

$(c)\;\; If $V_{n} = \int_{0}^{\frac{\pi}{2}}\frac{\sin^2(nx) }{\sin^2 x}dx\;,$ Where $n\in \mathbb{N}\;,$ Then $\sum^{100}_{r=1}V_{r} =$

Drsoccerball · Oct 22, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$Let $I = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx\;,$ Now put $x=\frac{1-t}{1+t} = -\left(\frac{t-1}{t+1}\right) = \frac{2}{1+t}-1$

$So we get $dx = -\frac{2}{(1+t)^2}dt\;,$ and changing limit\;, we get$

$I = \int_{1}^{0}\frac{\ln\left(\frac{2}{1+t}\right)}{2(1+t^2)}\cdot (1+t)^2\cdot \frac{-2}{(1+t)^2}dt = \int_{0}^{1}\left[\frac{\ln (2)-\ln(1+t)}{1+t^2}\right]dt$

$So we get $I = \ln (2)\int_{0}^{1}\frac{1}{1+t^2}dt - \int_{0}^{1}\frac{\ln (1+t)}{1+t^2}dt = \ln (2)\cdot \frac{\pi}{4}-\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx$

$So $I = \frac{\pi\cdot \ln(2)}{4}-I \Rightarrow 2I = \frac{\pi\cdot \ln(2)}{4}\Rightarrow I = \frac{\pi\cdot \ln(2)}{8}$

An alternative :

$$ let $ x= tan\theta$

$I=\int_0^{\frac{\pi}{4}}{ln(1+tan\theta)}d\theta$

$I= \int_0^{\frac{\pi}{4}}{ln(\frac{2}{1+tan\theta})}$

$2I= \int_0^{\frac{\pi}{4}}{ln(2)}$

$I=\frac{\pi ln(2)}{8}$

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