HSC 2017 MX2 Integration Marathon (archive) (1 Viewer)

stupid_girl · Feb 23, 2017

Re: HSC 2017 MX2 Integration Marathon

For Q4, if you stare at it long enough, you should be able to use reverse quotient rule.

Let y=1+ln x, then y'=1/x
The integrand can be written as
(y-1)/y^2
=(x'y-xy')/y^2.
Now, the answer should be obvious.

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
For Q4, if you stare at it long enough, you should be able to use reverse quotient rule.

Let y=1+ln x, then y'=1/x
The integrand can be written as
(y-1)/y^2
=(x'y-xy')/y^2.
Now, the answer should be obvious.

Also can work by $\times\frac{x}{x}$

jathu123 · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}}dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4.This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$

4.
$\int \frac{\ln x}{(1+\ln x)^2}$ d$x \\ = \int \frac{\ln x+1-\frac{x}{x}}{(1+\ln x)^2}$ d$x \\ =\frac{x}{1+\ln x}+$ C$$

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

jathu123 said:
4.
$\int \frac{\ln x}{(1+\ln x)^2}$ d$x \\ = \int \frac{\ln x+1-\frac{x}{x}}{(1+\ln x)^2}$ d$x \\ =\frac{x}{1+\ln x}+$ C$$

Correct!

Sorry if this seems like a dumb question, how do you go from line 2-3? Cause I can only do it your way by splitting the integral to $\int\frac{1}{1+lnx}dx-\int\frac{x}{x(1+lnx)^{2}}dx$ . Then if I do the 2nd integral by IBP the remaining integrals cancel, I get the answer.

jathu123 · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
Sorry if this seems like a dumb question, how do you go from line 2-3?

My bad, should have wrote it there. I just used reverse quotient rule, with the denominator being 1+lnx and the numerator x

Ohhh just realized stupid_girl already done this above

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

Wait what's this reserve quotient rule thing? I've never heard of this...

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
how do you go from line 2-3?

By inspection

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
Wait what's this reserve quotient rule thing? I've never heard of this...

$\noindent It's that $\int \frac{u'v -uv'}{v^2}\,\mathrm{d}x = \frac{u}{v} +c$ (reversing the quotient rule). jathu123 inspected that the integrand was of the form $\frac{u'v -uv'}{v^2}$.$

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

$\noindent Using similar reversals of rules you can come up with other rules, like the reverse product rule: $\int \left(u'v +uv'\rght)\,\mathrm{d}x = uv + c$.$

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
Are you sure Q3 has closed form solution?

$\text{For}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{re-write as}\int x (\frac{e^{x}}{1+e^{x}})dx\\\text{Then do IBP and a substitution in the remaining integral and bob's your uncle}$

leehuan · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

$\int_0^\pi t \prod_{\beta=1}^\infty \cos\left(\frac{t}{2^\beta}\right)dt$

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
For Q4, if you stare at it long enough, you should be able to use reverse quotient rule.

Let y=1+ln x, then y'=1/x
The integrand can be written as
(y-1)/y^2
=(x'y-xy')/y^2.
Now, the answer should be obvious.

OMG just saw it. That is legit awesome

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

leehuan said:
$\int_0^\pi \prod_{\beta=0}^\infty 2\cos\left(\frac{t}{2^\beta}\right)dt$

I think this will be divergent. It will be convergent though if you remove the 2 from inside the product.

leehuan · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

InteGrand said:
I think this will be divergent. It will be convergent though if you remove the 2 from inside the product.

I'm actually trying to remember what the question was. I can't recall it.

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

leehuan said:
I'm actually trying to remember what the question was. I can't recall it.

$\noindent What question? When I say remove 2, something like this will converge: $\prod_{k=0}^{\infty} \cos\left(\frac{t}{2^{k}}\right)$. You can then integrate that (as in it will be integrable). It can be shown that that product converges to $\frac{\sin t \cos t}{t}$. To make the resulting integral easier (expressible in terms of use of elementary functions), you could place a $t$ in front of the product to cancel the $t$ from the denominator of the result of the infinite product $\Bigg{(}$i.e. $t \prod_{k=0}^{\infty}\cos \left( \frac{t}{2^{k}} \right)$$\Bigg{)}$.$

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

InteGrand said:
$\noindent It's that $\int \frac{u'v -uv'}{v^2}\,\mathrm{d}x = \frac{u}{v} +c$ (reversing the quotient rule). jathu123 inspected that the integrand was of the form $\frac{u'v -uv'}{v^2}$.$

So when should I remember to see if this works?

BenHowe · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

InteGrand said:
$\noindent Using similar reversals of rules you can come up with other rules, like the reverse product rule: $\int \left(u'v +uv'\rght)\,\mathrm{d}x = uv + c$.$

How do I know to look for this in a question? It's a lot less intuitive than the quotient rule since there is no squared term on the denominator

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
How do I know to look for this in a question? It's a lot less intuitive than the quotient rule since there is no squared term on the denominator

It's basically just inspection. If you see some sum of product of functions like that, it might be a reverse product rule.

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
So when should I remember to see if this works?

If you see a square in the denominator.

InteGrand · Feb 24, 2017

Re: HSC 2017 MX2 Integration Marathon

leehuan said:
$\int_0^\pi \prod_{\beta=0}^\infty \cos\left(\frac{t}{2^\beta}\right)dt$

Yep this is now integrable, but I don't think you can find a closed form answer for it using typical high school functions / concepts only (can do with the Sine integral, just by definition of the Sine integral. The curious student can read more about such functions here: https://en.wikipedia.org/wiki/Trigonometric_integral .).

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