HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

snail489 · Oct 3, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int{\frac{(a\sin{x})^2 + (b\cos{x})^2}{(a^2 \sin{x})^2+(b^2\cos{x})^2}}dx$

$a,b$ are real non-zero constants

Should look like this:

Did this by dividing through b^4 tan^4x and then splitting the numerator into two fractions. Then substituting u^(1/3) = (a/b)tanx and w=(a/b)tanx. Then it simplifies into algebraic integrals like 1/(w^4+1) which can be evaluated using partial fractions. Although I think I probably did some unnecessary steps (as per usual) since half my answer cancelled out at the final step.

juantheron · Oct 4, 2015

Re: MX2 2015 Integration Marathon

$Let $I = \int\frac{1}{\sin^ 6+\cos^6 x}dx = \int\frac{1}{1-3\sin^2 x\cdot \cos^2 x}dx = \int\frac{4}{4-3\sin^2 (2x)}dx$

$Now\; Divide\; both\; Numerator\; and \; Denominator\; by\; $\cos^2(2x),$ We\; get$

$I = \int\frac{4\sec^2 2x}{4(1+\tan^2 2x)-3}dx = \int\frac{\sec^2(2x)}{\tan^2(2x)+\frac{1}{4}}dx$

$Now\; Put \; $\tan (2x) = t\;,$ Then $2\sec^2 (2x)dx = dt$\;, We\; Get$

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

juantheron · Oct 4, 2015

Re: MX2 2015 Integration Marathon

$Let $I = \int\frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx$\;, Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^2 x.$

$So\; we\; get $I = \int\frac{a^2 \tan^2 x+b^2}{a^4\tan^2 x+b^4}dx$

$Now\; Put \; $\tan x=t\;,$ Then $\sec^2 dx = dt$

$I = \int\frac{a^2t^2+b^2}{(a^4t^2+b^2)\cdot (1+t^2)}dt = \frac{1}{a^2+b^2}\cdot \int \left[\frac{1}{t^2+1}+\frac{a^2b^2}{a^4t^2+b^4}\right]dx$

$I = \frac{1}{a^2+b^2}\int\frac{1}{t^2+1}dt+\frac{b^2}{a^2(a^2+b^2)}\cdot \int\frac{1}{t^2+\left(\frac{b^2}{a^2}\right)^2}dt$

$I = \frac{1}{a^2+b^2}\cdot \tan^{-1}(t)+\frac{b^2}{a^2(a^2+b^2)}\cdot \frac{a^2}{b^2}\tan^{-1}\left(\frac{a^2t}{b^2}\right)+\mathcal{C}$

$I = \frac{1}{a^2+b^2}\cdot \tan^{-1}(\tan x)+\frac{1}{a^2+b^2}\cdot \tan^{-1}\left(\frac{a^2\tan x}{b^2}\right)+\mathcal{C}$

juantheron · Oct 4, 2015

Re: MX2 2015 Integration Marathon

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

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

leehuan · Oct 4, 2015

Re: MX2 2015 Integration Marathon

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seanieg89 · Oct 4, 2015

Re: MX2 2015 Integration Marathon

For any who liked the BoS trials question about the trapezoidal rule, the linked article explores those kind of approximations more.

It's an interesting application of HSC integration theory and not enough teachers talk about these things imo.

http://community.boredofstudies.org/14/mathematics-extension-2/342900/numerical-integration.html

leehuan · Oct 4, 2015

Re: MX2 2015 Integration Marathon

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Paradoxica · Oct 4, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
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WOOO!!!!
IBP 7 times!

InteGrand · Oct 4, 2015

Re: MX2 2015 Integration Marathon

$Alternatively, let $\mathcal{J}_k = \binom{207}{k}\mathcal{I}_k$, where $\mathcal{I}_k = \int _0 ^1 x^{207 -k}(1-x)^k\text{ d}x$, for $k=0,1,...,207$. We want to find $\mathcal{J}_7$. By IBP, we find$

$\mathcal{I}_k = 0 + \int _0 ^1 \frac{x^{207 - (k-1)}}{207 - k+1}\cdot k(1-x)^{k-1}\text{ d}x$

$$= \frac{k}{207-k+1}I_{k-1}$, $k\geq 1$.$

$So $\mathcal{J}_k = \binom{207}{k}\mathcal{I}_k$$

$= \frac{207!}{k! (207-k)!}\times \frac{k}{201-k+1}I_{k-1}$

$= \frac{207!}{(k-1)! (207-k+1)!}I_{k-1}$

$= \frac{207!}{(k-1)! (207-(k-1))!}I_{k-1}$

$$= \binom{207}{k-1} I_{k-1}$.$

$$= \mathcal{J}_{k-1}$ for all $k\geq 1$, so the $\mathcal{J}_k$'s are a constant $\mathcal{J}$ for $k=0,1,...,207$.$

$But $\sum_{k=0} ^n \mathcal{J}_k =\sum_{k=0} ^n \int _0 ^1 \binom{207}{k} x^{207 -k}(1-x)^k\text{ d}x$$

$=\int _0 ^1 \sum_{k=0} ^n \binom{207}{k} x^{207 -k}(1-x)^k\text{ d}x$

$$=\int _0 ^1 1\text{ d}x$ (by Binomial theorem)$

$$=1$.$

$Since this sum consists of $208$ constant terms $\mathcal{J}$, it follows that $\mathcal{J} = \frac{1}{208}$ (answer).$

$(Or more simply, the constant $\mathcal{J}$ is clearly the value of $\mathcal{J}_0$, from the recurrence relation $\mathcal{J}_k = \mathcal{J}_{k-1}$, so $\mathcal{J}=\mathcal{J}_0 = \int _0 ^1 x^{207}\text{ d}x = \frac{1}{208}.$)$

Drsoccerball · Oct 4, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
WOOO!!!!
IBP 7 times!

You could also create a recurrence formula making it easier.

Carrotsticks · Oct 4, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
WOOO!!!!
IBP 7 times!

Tabular IBP takes care of it nicely without it really seeming like 7 applications of IBP.

Carrotsticks · Oct 4, 2015

Re: MX2 2015 Integration Marathon

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

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

Drsoccerball · Oct 4, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Tabular IBP takes care of it nicely without it really seeming like 7 applications of IBP.

Doesn't tabular integration only work when the function is differentiated infinite times and its still not 0 ? How can we use that if its 0 eventually?

Carrotsticks · Oct 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Doesn't tabular integration only work when the function is differentiated infinite times and its still not 0 ? How can we use that if its 0 eventually?

No?

Tabular integration by parts is mechanically identical to repeated applications of integration by parts. It just arranges things in a neater manner (the table) that makes the iterations a lot cleaner to work with.

InteGrand · Oct 4, 2015

Re: MX2 2015 Integration Marathon

Tabular integration was popularised in the movie Stand and Deliver.

leehuan · Oct 6, 2015

Re: MX2 2015 Integration Marathon

Assuming the n should be an r or vice versa, any hint on the reduction formula? I got scared doing sin(n-1)xcos(x)+cos(n-1)xsin(x)

InteGrand · Oct 11, 2015

Re: MX2 2015 Integration Marathon

$Let $a<b$ be real numbers and let $f$ be any real-valued function defined on $[a,b]$ such that $\int _a ^b f(x)\text{ d}x$ exists.$

$(i) Show that $\int _a ^b f(x)\text{ d}x = \int _a ^b f(a+b-x)\text{ d}x$.$

$(ii) Describe the geometrical significance of the result in (i).$

kawaiipotato · Oct 11, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $a<b$ be real numbers and let $f$ be any real-valued function defined on $[a,b]$ such that $\int _a ^b f(x)\text{ d}x$ exists.$

$(i) Show that $\int _a ^b f(x)\text{ d}x = \int _a ^b f(a+b-x)\text{ d}x$.$

$(ii) Describe the geometrical significance of the result in (i).$

$$ (i) $ $ let $ u = a+b-x$

$du = -dx$

$When x = a, u = b$

$When x = b, u = a$

$\therefore $ I $ = - \int_{b}^{a} f(a+b - u) du$

$= \int_{a}^{b} f(a+b-u) du$

$= \int_{a}^{b} f(a+b-x) dx ($dummy variables$)$

$(ii) f(a+b-x) $ is the $ f(x) $ curve mirrored against the $ x = \frac{a+b}{2}$ curve $$

$$ Hence, $ \int _a ^b f(x)\text{ d}x = \int _a ^b f(a+b-x)\text{ d}x $means that the area bounded by f(x) in [a,b] equals to the area of f(x) ,mirrored against the $ x = \frac{a+b}{2} $ line $ $in $ [a,b]$

Drongoski · Oct 11, 2015

Re: MX2 2015 Integration Marathon

Another way.

Say:

$I = \int ^b _a f(x) dx = F(b) - F(a)\\ \\ \int ^b _a f(a+b-x)dx = - \int ^b _4 f(a+b-x) d(a+b-x) \\ \\ =- \left [ F(a+b-x) \right]^b _a \\ \\ = - [ F(a+b - b) - F(a+b - a)] \\ \\ = -[F(a) - F(b)] \\ \\ = F(b) - F(a) \\ \\ = I$

The geometrical meaning: f(a+b-x) is mirror image of f(x) on the y-axis, shifted right by a+b units. It can be shown that the area associated with I is the same area as the new definite integral.

Paradoxica · Oct 11, 2015

Re: MX2 2015 Integration Marathon

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

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