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

glittergal96 · Nov 17, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Classic trick to integrating the inverse trig functions. Put it in your toolbox!

Or more generally the inverse functions of ANYTHING we already know how to integrate

.

omegadot · Nov 18, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
NEW QUESTION:
$\int\sin{(\ln{x})}\,dx$

$$\noindent Use IBP twice. Doing so we have$\\\begin{align*}\int \sin (\ln x) \, dx &= x \sin (\ln x) - \int \cos (\ln x) \, dx\\&=x \sin (\ln x) - x \cos (ln x) - \int \sin (\ln x) \, dx\\\Rightarrow 2 \int \sin (\ln x) \, dx &= x \sin (\ln x) - x \cos (\ln x)\\\Rightarrow \int \sin (\ln x) \, dx &= \frac{x}{2} \left [\sin (\ln x) - \cos (\ln x) \right ] + \cal{C}\end{align*}$

$\noindent \textbf{\textsc{Next Question}}$

$$\noindent One for the 2016ers. Find $ \int \frac{dx}{x^4 + 2016 x^2 + 1}.$

leehuan · Nov 18, 2015

Re: MX2 2016 Integration Marathon

Whilst the above question lies there for the 2016ers (which.. tbh whilst it's not hard the number 2016 will scare people off)

Meanwhile, for the others:

$\\ \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2008 }{ x } }{ \left( { 2007 }^{ x }+1 \right) \left( \sin ^{ 2008 }{ x } +\cos ^{ 2008 }{ x } \right) } dx }$

Paradoxica · Nov 18, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent One for the 2016ers. Find $ \int \frac{dx}{x^4 + 2016 x^2 + 1}.$

$Complete the square.$

$I = \int \frac{dx}{(x^2+ 1008)^2 - \alpha^2} $ where $\alpha = 2016^2 -1$

$Partial Fractions by algebraic manipulation.$

$I = \frac{1}{2\alpha} \int \frac{\alpha + \alpha}{(x^2+ 1008)^2 - \alpha^2}dx = \frac{1}{2\alpha} \int \frac{x^2 + 1008 + \alpha -( x^2 + 1008 - \alpha)}{(x^2+ 1008 + \alpha)(x^2 + 1008 - \alpha)}dx$

$I = \frac{1}{2\alpha} \int \frac{1}{x^2 + 1008 - \alpha} -\frac{1}{x^2+ 1008 + \alpha} dx $ Note that as both denominators are strictly positive, we only need the inverse tangent integral formula to bash this integral. $$

$The remainder of this integral is left as an exercise to the students.$

leehuan said:
Whilst the above question lies there for the 2016ers (which.. tbh whilst it's not hard the number 2016 will scare people off)

Meanwhile, for the others:

$\\ \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2008 }{ x } dx }{ \left( { 2007 }^{ x }+1 \right) \left( \sin ^{ 2008 }{ x } +\cos ^{ 2008 }{ x } \right) } }$

$$Use the border-flip substitution $s = a+b-x $ where $a$ and $b$ are the limits of integration.$

$I = \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2008 }{ x } dx }{ \left( { 2007 }^{ x }+1 \right) \left( \sin ^{ 2008 }{ x } +\cos ^{ 2008 }{ x } \right) } } = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2007^x \sin^{2008}{x}(-dx)}{(2007^x + 1)(\sin^{2008}{x}+\cos^{2008}{x})}$

$The negative on the $dx$ unflips the borders of integration, allowing us to add the two identical definite integrals together.$

$2I = \int_{-\frac { \pi }{ 2 }}^{\frac { \pi }{ 2 }}\frac{(2007^x + 1)(\sin^{2008}{x})dx}{(2007^x +1)(\sin^{2008}{x}+\cos^{2008}{x})} = \int_{-\frac { \pi }{ 2 }}^{\frac { \pi }{ 2 }}\frac{\sin^{2008}{x}dx}{\sin^{2008}{x}+\cos^{2008}{x}}$

$2I = 2 \int_{0}^{\frac { \pi }{ 2 }}\frac{\sin^{2008}{x}dx}{\sin^{2008}{x}+\cos^{2008}{x}} $ Inciting the property of definite integrals of even functions$$

$The rest can be done using the standard degenerate border flip substitution.$

omegadot · Nov 19, 2015

Re: MX2 2016 Integration Marathon

$$\noindent Next question for the 2016ers. Find $\int \frac{dx}{x - x^{2016}}.$

Paradoxica · Nov 21, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Next question for the 2016ers. Find $\int \frac{dx}{x - x^{2016}}.$

$Factorise $x^{2016}$out of the denominator and use reverse chain rule$

leehuan · Nov 21, 2015

Re: MX2 2016 Integration Marathon

Easy question to make sure the 2016ers don't get scared away

$\\ \int { \frac { dx }{ x\sqrt { { x }^{ 2 }-1 } } }$

omegadot · Nov 21, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
$Factorise $x^{2016}$out of the denominator and use reverse chain rule$

$$\noindent Correct. For the benefit of others, from Paradoxia's suggestion we have$

$\begin{align*}\int \frac{dx}{x - x^{2016}} &= \int \frac{dx}{x^{2016}(x^{-2015} - 1)}\\&= \int \frac{x^{-2016}}{x^{-2015} - 1} dx.\end{align*}$

$\noindent Let $u = x^{-2015} \Rightarrow du = -2015 x^{-2016} \, dx$. So the integral becomes$

$\begin{align*}\int \frac{dx}{x - x^{2016}} &= -\frac{1}{2015} \int \frac{du}{u - 1}\\&= -\frac{1}{2015} \ln |u - 1| + {\cal{C}}\\&= -\frac{1}{2015} \ln \left |\frac{1}{x^{2015}} - 1 \right | + \cal{C}.\end{align*}$

$\noindent \textbf{Next Question}$

$$\indent If $a \in \mathbb{R}$ find $\int \frac{\sin (x - a)}{\sin (x + a)} dx.$

Drsoccerball · Nov 21, 2015

Re: MX2 2016 Integration Marathon

I don't think Paradoxica can be considered as a 2016'er with integration

...

leehuan · Nov 22, 2015

Re: MX2 2016 Integration Marathon

$\int { \frac { \sin { \left( \left( x+a \right) -2a \right) } }{ \sin { \left( x+a \right) } } dx } \\ =\int { \frac { \sin { \left( x+a \right) } \cos { 2a } -\cos { \left( x+a \right) \sin { 2a } } }{ \sin { \left( x+a \right) } } dx } \\ =\cos { 2a } \int { dx } -\sin { 2a } \int { \frac { \cos { \left( x+a \right) } }{ \sin { \left( x+a \right) } } dx } \\ =x\cos { 2a } -\sin { 2a } \ln { \left| \sin { \left( x+a \right) } \right| } +C$

Paradoxica · Nov 22, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I don't think Paradoxica can be considered as a 2016'er with integration ...

I see how it is. >: D

porcupinetree · Nov 22, 2015

Re: MX2 2016 Integration Marathon

Next question:

$\int \frac{x^{2} - 7x + 3}{\sqrt{x^{2} + 2x + 7}} dx$

leehuan · Nov 22, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
Next question:

$\int \frac{x^{2} - 7x + 3}{\sqrt{x^{2} + 2x + 7}} dx$

I think the integral of 1/sqrt(x^2+a^2) has been taken off the formula sheet.

porcupinetree · Nov 22, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I think the integral of 1/sqrt(x^2+a^2) has been taken off the formula sheet.

:O you're right, I just checked

(but for now let's just pretend that they didn't take it off)

InteGrand · Nov 22, 2015

Re: MX2 2016 Integration Marathon

For those who want to treat it as though that integral is still given, it is this:

$$$\int \frac{1}{\sqrt{ x^2+a^2 }} \text{ d}x =\ln \left(x + \sqrt{x^2 + a^2} \right) + C$

.

omegadot · Nov 22, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Next question for the 2016ers. Find $\int \frac{dx}{x - x^{2016}}.$

$$\noindent Alternatively, if one recognises$\\\begin{align*}\frac{1}{x - x^{2016}} = \frac{1}{x(1 - x^{2015})} = \frac{1}{x} + \frac{x^{2014}}{1 - x^{2015}},\end{align*}$

$$\noindent One has\\\begin{align*}\int \frac{dx}{x - x^{2016}} &= \int \frac{dx}{x} + \int \frac{x^{2014}}{1 - x^{2015}}\\&= \int \frac{dx}{x} - \frac{1}{2015} \int \frac{-2105 x^{2014}}{1 - x^{2015}} dx\\&= \ln|x| - \frac{1}{2015} \ln \left |1 - x^{2015} \right | + {\cal{C}}\end{align*}$

leehuan · Nov 24, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
Next question:

$\int \frac{x^{2} - 7x + 3}{\sqrt{x^{2} + 2x + 7}} dx$

I started doing this, but the denominator is horrible... got lazy when one of my integrands was sqrt(x^2+2x+7) [in the numerator over 1]

porcupinetree · Nov 24, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I started doing this, but the denominator is horrible... got lazy when one of my integrands was sqrt(x^2+2x+7) [in the numerator over 1]

It is indeed quite tedious, but nonetheless a good exercise as it requires several techniques.
On another note, are there actually any 2016'ers who are participating in these maths marathons? (besides Paradoxica, he doesn't count

)

Paradoxica · Nov 24, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
It is indeed quite tedious, but nonetheless a good exercise as it requires several techniques.
On another note, are there actually any 2016'ers who are participating in these maths marathons? (besides Paradoxica, he doesn't count )

I see how it is. *glares*

leehuan · Nov 24, 2015

Re: MX2 2016 Integration Marathon

Get over it Paradox. When it comes to maths you're at latest a 15'er

-------------
Not sure if KingOfActing will go for it

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

Active Member

Active Member

Well-Known Member

-insert title here-

Active Member

-insert title here-

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

-insert title here-

not actually a porcupine

Well-Known Member

not actually a porcupine

Well-Known Member

Active Member

Well-Known Member

not actually a porcupine

-insert title here-

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 5)