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

mreditor16 · Jun 17, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Really? By $D^\prime(x)$ , I meant the derivative of $D(x)$ . So in the second quoted post, basically only one "tedious" step is required, which is to solve a linear system of equations in two unknowns.

Once we do this, we're pretty much done, because the integral of the $\frac{D^\prime (x)}{D(x)}$ part is just $\ln |D(x)|$ .

i.e. $3\sin x + 2\cos x \equiv A(2\sin x + 3\cos x)+B(-3\sin x + 2\cos x)$ , where $A=\frac{12}{13},B=-\frac{5}{13}$ (assuming I didn't make a silly mistake in solving the simultaneous equations, but you get the idea).

Then

$\int \frac{3\sin x + 2\cos x}{2\sin x + 3\cos x}\text{ d}x=\int \frac{A(2\sin x + 3\cos x)+B(-3\sin x + 2\cos x)}{2\sin x + 3\cos x}\text{ d}x$

$=\int \left(A+B\frac{-3\sin x + 2\cos x}{2\sin x + 3\cos x} \right)\text{ d}x$

$=Ax+B\ln |2\sin x + 3\cos x|+c$ .

I'd say that's not too messy.

Oh yeh it comes out nicely, I think I might have made some mistake along the way :/ But yeah that comes out neatly - thanks heaps Integrand

Drsoccerball · Jun 17, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { 1+\sin { x } +\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } }{ 1+\cos { x } +\cos ^{ 2 }{ x } +\cos ^{ 3 }{ x } } dx }$

I must resort to t substitution

Sy123 · Jun 18, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { 1+\sin { x } +\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } }{ 1+\cos { x } +\cos ^{ 2 }{ x } +\cos ^{ 3 }{ x } } dx }$

$\\ $See$ \ f(z) = \frac{2 - z^2}{1 + z + z^2 + z^3} \\ \\ = \frac{(2-z^2)(1-z)}{1 - z^4} = \frac{1}{2} \cdot \frac{1}{1+z} - \frac{3}{2} \cdot \frac{z}{1+z^2} + \frac{3}{2} \cdot \frac{1}{1+z^2}$

$\\ I = \int { \frac { 1+\sin { x } +\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } }{ 1+\cos { x } +\cos ^{ 2 }{ x } +\cos ^{ 3 }{ x } } dx } \\ \\ = \int f(\cos x) + \sin x f(\cos x) \ dx = \int f(\cos x) \ dx + \int \sin x f(\cos x) \ dx = I_1 + I_2$

$\\ I_2 = \int \sin x f(\cos x) \ dx, \ u = \cos x$

$\\ I_2 = - \int f(u) \ du , \\ $(which is easily integratable by seeing the partial fraction form given above)$$

$\\ I_1 = \int f(\cos x) \ dx$

$\\ $This integral involves the integration of: $ \\ J = \int \frac{1}{1+ \cos x} \ dx = \int \csc x^2 - \cot x \ dx = - \cot x + \csc x + c_j \\ \\ K = \int \frac{\cos x}{1 + \cos^2x} \ dx = \int \frac{\cos x}{2 - \sin^2 x} , \ v = \sin x \\ \Rightarrow K = \frac{1}{2\sqrt{2}} \ln \left|\frac{\sqrt{2} + \sin x}{\sqrt{2} - \sin x} \right|+ c_k \\ \\ L = \int \frac{1}{1 + \cos^2x} \ dx = \int \frac{\csc^2x}{2\cot^2x + 1} \ dx , w = \cot x \\ \Rightarrow L = \frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2} \cot x) + c_L$

Then the final answer is just various combinations of all these integrals

Is there an easier way?

leehuan · Jun 20, 2015

Re: MX2 2015 Integration Marathon

$$1. $\int { \frac { 2015-{ x }^{ n } }{ 1-x } dx }$

$$2. $\int { { e }^{ x }\sqrt { 1+{ e }^{ 2x } } dx }$

$$3. $\int _{ 0 }^{ 1 }{ \frac { 1-x }{ \left( 1+x \right) \ln { x } } dx }$

Drsoccerball · Jun 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$$1. $\int { \frac { 2015-{ x }^{ n } }{ 1-x } dx }$

$\int{\frac{2015}{1-x} +\frac{x^n}{x-1}}dx$

$-2015ln|1-x| + \int{\frac{x^n-1+1}{x-1}}$

$-2015ln|1-x|+ln|x-1| +\int{\frac{(x-1)(x^{n-1} +...+x+1)}{x-1}}$

$\therefore = -2015ln|1-x| +ln|x-1| + \frac{x^n}{n}+\frac{x^{n-1}}{n-1}+...+ \frac{x^2}{2} +x +C$

Drsoccerball · Jun 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$$2. $\int { { e }^{ x }\sqrt { 1+{ e }^{ 2x } } dx }$

$let x=ln|tan\theta |$

$\int{sec^3\theta}d\theta = \int{e^x\sqrt{1+e^{2x}}}dx$

$\int{sec^3\theta}dx = \int{sec\theta (1+tan^2\theta)}$

$\int{sec \theta} = ln|sec \theta +tan \theta |$

$\int{sec \theta tan^2\theta} = sec^2\theta - \int{tan\theta +tan^3\theta} $ By IBP $$

$\int{secxtan^2x} = sec^2\theta +ln|cos\theta | - \int{tan^3\theta}$

$- \int{tan^3\theta} = - \frac{tan^2\theta}{2} + ln|sec^2\theta|$

$\therefore \int{sec^3\theta}= sec^2\theta -\frac{tan^2\theta}{2} +ln|sec\theta + tan\theta| + ln|sec\theta| +C$

$tan^{-1}(e^x) = \theta $ Sub all theta $$

[]

leehuan · Jun 20, 2015

Re: MX2 2015 Integration Marathon

For 1 you could've just done...

$\int { \left( \frac { 2014 }{ 1-x } +\frac { 1-{ x }^{ n } }{ 1-x } \right) dx } \\ =-2014\log { \left| 1-x \right| } +\int { \left( 1+x+{ x }^{ 2 }+\dots +{ x }^{ n-1 } \right) dx } \\ =-2014\log { \left| 1-x \right| } +x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\dots +\frac { { x }^{ n } }{ n } +C$

Drsoccerball · Jun 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
For 1 you could've just done...

$\int { \left( \frac { 2014 }{ 1-x } +\frac { 1-{ x }^{ n } }{ 1-x } \right) dx } \\ =-2014\log { \left| 1-x \right| } +\int { \left( 1+x+{ x }^{ 2 }+\dots +{ x }^{ n-1 } \right) dx } \\ =-2014\log { \left| 1-x \right| } +x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\dots +\frac { { x }^{ n } }{ n } +C$

One step isnt that much of a big deal

how do you do 3 ?

leehuan · Jun 20, 2015

Re: MX2 2015 Integration Marathon

I know

I had a long think about Q3. The hint the sheet gave was to use log laws but the fact that there is no standard result, it might not be suitable for Ext 2 maths anyhow... Don't see how the f(a-x) - f(x) trick will work either. I'll leave that question unless one of the experts can take care of it.

Drsoccerball · Jun 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
I know

I had a long think about Q3. The hint the sheet gave was to use log laws but the fact that there is no standard result, it might not be suitable for Ext 2 maths anyhow... Don't see how the f(a-x) - f(x) trick will work either. I'll leave that question unless one of the experts can take care of it.

Could you do it doe?

leehuan · Jun 22, 2015

Re: MX2 2015 Integration Marathon

Nope. I could only do this.

Screen Shot 2015-06-22 at 10.34.00 am.png

$\int { \cot ^{ -1 }{ \left( { x }^{ 2 }+x+1 \right) } dx }$

Note: $\cot ^{ -1 }{ x } =\tan ^{ -1 }{ \frac { 1 }{ x } }$

Possibly useful compound angle formula: $\cot { \left( a+b \right) } =\frac { \cot { a } \cot { b } -1 }{ \cot { a } +\cot { b } }$

Drsoccerball · Jun 22, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Nope. I could only do this.
View attachment 32134

$\int { \cot ^{ -1 }{ \left( { x }^{ 2 }+x+1 \right) } dx }$

Note: $\cot ^{ -1 }{ x } =\tan ^{ -1 }{ \frac { 1 }{ x } }$

Possibly useful compound angle formula: $\cot { \left( a+b \right) } =\frac { \cot { a } \cot { b } -1 }{ \cot { a } +\cot { b } }$

Btw in wolfram alpha you can use latex to type: eg: int_0^1((sinx)/(x))

porcupinetree · Jun 29, 2015

Re: MX2 2015 Integration Marathon

Next integral:

sqrt (1 + sin2x) dx

RecklessRick · Jun 29, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Next integral:

sqrt (1 + sin2x) dx

Idk how to LaTeX (can someone point me to a guide of some sort?) so here goes:

sqrt(1+sin2x)
= sqrt(1 + 2sinxcosx)
= sqrt(sin^2x + 2sinxcosx + cos^2x)
= sqrt((sinx + cosx)^2)
= sinx + cosx

so integral sqrt(1 + sin2x) dx
= sinx - cosx + c

kawaiipotato · Jun 29, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Next integral:

sqrt (1 + sin2x) dx

= sqrt(sin^2 x + cos^2 x + 2sinxcosx) dx

= sqrt((sinx + cosx)^2) dx
= |sinx+cosx|

porcupinetree · Jun 29, 2015

Re: MX2 2015 Integration Marathon

Next question:

sqrt [(5-x)/(x-1)] dx

leehuan · Jun 29, 2015

Re: MX2 2015 Integration Marathon

$$If we can assume $x\neq 5\\ \int { \sqrt { \frac { 5-x }{ x-1 } } dx } \\ =\int { \frac { 5-x }{ \sqrt { -{ x }^{ 2 }+6x-5 } } dx } \\ =\int { \frac { 2 }{ \sqrt { 4-{ \left( x-3 \right) }^{ 2 } } } dx } +\int { \frac { -x+3 }{ \sqrt { -{ x }^{ 2 }+6x-5 } } dx } \\ =2\arcsin { \frac { x-3 }{ 2 } } +\sqrt { -{ x }^{ 2 }+6x-5 } +C$

$$Alternatively, let $x=\sin ^{ 2 }{ \theta }$

leehuan said:
$\int { \cot ^{ -1 }{ \left( { x }^{ 2 }+x+1 \right) } dx }$

Note: $\cot ^{ -1 }{ x } =\tan ^{ -1 }{ \frac { 1 }{ x } }$

Possibly useful compound angle formula: $\cot { \left( a+b \right) } =\frac { \cot { a } \cot { b } -1 }{ \cot { a } +\cot { b } }$

kawaiipotato · Jun 29, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \cot ^{ -1 }{ \left( { x }^{ 2 }+x+1 \right) } dx }$

$$ From the identity, $ cot(a+b) = \frac{cotacotb - 1}{cota+cotb}$

$cot^{-1}a - cot^{-1}b = cot^{-1}(\frac{ab - 1}{a+b})$

$$ I = $ \int cot^{-1}(x(x+1) - 1) dx$

$$ I = $ \int cot^{-1}(\frac{x(x+1) - 1}{(x+1) - (x)}) dx$

$$ I = $ \int cot^{-1}(x+1) - cot^{-1}x dx$

$$ I = $ \int tan^{-1}{\frac{1}{x+1}} - tan^{-1}(\frac{1}{x}}) dx$
$Integrate by parts$

leehuan · Jun 29, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$$ From the identity, $ cot(a+b) = \frac{cotacotb - 1}{cota+cotb}$

$cot^{-1}a - cot^{-1}b = cot^{-1}(\frac{ab - 1}{a+b})$

$$ I = $ \int cot^{-1}(x(x+1) - 1) dx$

$$ I = $ \int cot^{-1}(\frac{x(x+1) - 1}{(x+1) - (x)}) dx$

$$ I = $ \int cot^{-1}(x+1) - cot^{-1}x dx$

Instead of applying the 'not spoilered' identity immediately I would do this:

$\\ \frac { d }{ dx } \cot ^{ -1 }{ x } \\ =\frac { d }{ dx } \tan ^{ -1 }{ \frac { 1 }{ x } } \\ =\frac { 1 }{ 1+\frac { 1 }{ { x }^{ 2 } } } \cdot -\frac { 1 }{ { x }^{ 2 } } \\ =-\frac { 1 }{ { x }^{ 2 }+1 }$

So that I can use integration by parts at where I cut off your quote.

NEXT QUESTION:
$\int { { x }^{ 8 }\sin { \left( { x }^{ 3 } \right) } \cos { \left( { x }^{ 3 } \right) } \tan { \left( { x }^{ 3 } \right) } dx }$

Ekman · Jun 29, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Next question:

sqrt [(5-x)/(x-1)] dx

Alternate method:

$$Let: $ x=u+3$

$\therefore \int \sqrt{\frac{5-x}{x-1}} dx = \int \sqrt{\frac{2-u}{2+u}} du = \int \frac{2-u}{\sqrt{4-u^2}} du$

$= 2\sin^{-1}{\frac{u}{2}} +\sqrt{4-u^2} +c$

$\therefore 2\sin^{-1}{\frac{x+3}{2}} +\sqrt{4-(x+3)^2} +c$

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