HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

braintic · Jan 10, 2014

Re: MX2 Integration Marathon

$\int \ln x \cdot \sin^{-1} x \ dx$

dunjaaa · Jan 27, 2014

Re: MX2 Integration Marathon

Finally, I managed to get this one out, which reminds me of a similar u-sub question you posted in the 2013 3u marathon

dunjaaa · Jan 27, 2014

Re: MX2 Integration Marathon

Just learnt integration by parts today, man that's some grueling algebra haha

.

asianese · Jan 27, 2014

Re: MX2 Integration Marathon

You won't get anything like those in the HSC mind you!

Carrotsticks · Jan 27, 2014

Re: MX2 Integration Marathon

asianese said:
You won't get anything like those in the HSC mind you!

='(

I enjoy giving a class a difficult integration problem, and observing what methods are used to solve it. It's so interesting! Some people think of the craziest ways, some people think of the most brute-force ways, some people think of elegant ways etc.

Sy123 · Jan 27, 2014

Re: MX2 Integration Marathon

This is always a good one

$\int_0^{\pi /2} x \cot x \ dx$

asianese · Jan 27, 2014

Re: MX2 Integration Marathon

Sy123 said:
This is always a good one

$\int_0^{\pi /2} x \cot x \ dx$

I remember this from 2012 hahha. Leme see if I remember how to do it... (ceebs actually)

Sy123 · Jan 27, 2014

Re: MX2 Integration Marathon

$\int \frac{\sin x}{\sqrt{1+ \sin x}} \ dx$

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

Sy123 said:
This is always a good one

$$Let: I $= \int_0^{\pi /2} x \cot x \ dx$

$I = \int_0^{\frac{\pi}{2}} \frac{xcosx \ dx}{sinx} \\ \\ \\ $Let u=x$,$ $ dv=\frac{cosx \ dx}{sinx} \\ \\ \\ du=dx,$ $ v=ln(sinx) \\ \\ \\ I= uv - \int vdu \\ \\ \\ I= [x \ ln(sinx)]^{\frac{\pi}{2}}_{0} - \int_0^{\frac{\pi}{2}} ln(sinx) \ dx \\ \\ \\ $Let $u=ln(sinx), \ dv=dx \\ \\ \\ du= \frac{cosx \ dx}{sinx}, v=x \\ \\ \\ I= 0 - [x \ ln(sinx)]^{\frac{\pi}{2}}_{0} + I$

Well at least I know all my working out is correct. As for the method, not quite.

Second attempt:

$I = \int_0^{\frac{\pi}{2}} \frac{xcosx \ sinx \ dx}{sin^{2}x} \\ \\ \\ $Let u=x, $ dv=\frac{sinx \ cosx}{sin^{2}x} \\ \\ \\ du=dx, $ $ v=\frac{ln(sinx)}{2} \\ \\ \\ I= [x \frac{ln(sinx)}{2}]^{\frac{\pi}{2}}_{0} - \int_0^{\frac{\pi}{2}} \frac{ln(sinx) \ dx}{2} \\ \\ \\ Let $ $ u= \frac{ln(sinx)}{2},$ $ dv = dx \\ \\ \\ du= \frac{cosx \ dx}{2sinx},$ $ v=x \\ \\ \\ I= 0 - \frac{x \ ln(sinx)}{2} + \int_0^{\frac{\pi}{2}} \frac{x \ cosx \ dx}{2sinx} \\ \\ \\ 2I = I \\ \\ \\ \therefore{I=0}$

Something is wrong with the working out here. I probably need to substitute instead. I'll give it one more go:

$Let: x= \frac{\pi}{2} - u. \\ \\ \\ I= -\int^{0}_{\frac{\pi}{2}} \frac{ (\frac{\pi}{2} - u) \ cos(\frac{\pi}{2} - u) \ du}{sin(\frac{\pi}{2} - u)} \\ \\ \\ $Using what I had above through integration by parts: $ \\ \\ \\ I= [(\frac{\pi}{2})(ln({sin(\frac{\pi}{2} - u)}))]_{0}^{\frac{\pi}{2}} - [u \ ln(sin(\frac{\pi}{2} - u))]^{\frac{\pi}{2}}_{0} + \int_0^{\frac{\pi}{2}} ln(sin(\frac{\pi}{2} - u)) \\ \\ \\ I =[(\frac{\pi}{2})(ln({sin(\frac{\pi}{2} - u)}))]_{0}^{\frac{\pi}{2}} - [u \ ln(sin(\frac{\pi}{2} - u))]^{\frac{\pi}{2}}_{0} + [u \ ln(sin(\frac{\pi}{2}-u))]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} ln({sin(\frac{\pi}{2}}-u))} du \\ \\ \\ I= [(\frac{\pi}{2})(ln({sin(\frac{\pi}{2} - u)}))]_{0}^{\frac{\pi}{2}} - [u \ ln(sin(\frac{\pi}{2} - u))]^{\frac{\pi}{2}}_{0} + [u \ ln(sin(\frac{\pi}{2}-u))]_0^{\frac{\pi}{2}} -I$

$\therefore I = \frac{[(\frac{\pi}{2})(ln({sin(\frac{\pi}{2} - u)}))]_{0}^{\frac{\pi}{2}} - [u \ ln(sin(\frac{\pi}{2} - u))]^{\frac{\pi}{2}}_{0} + [u \ ln(sin(\frac{\pi}{2}-u))]_0^{\frac{\pi}{2}}}{2}$

Its too messy for me now on latex. Also, the first part is completely incorrect as it gives me a tendency to negative infinity.

HeroicPandas · Jan 27, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{\sin x}{\sqrt{1+ \sin x}} \ dx$

$$Let $I = \int \frac{sinx}{\sqrt{1+sinx}}dx \\ \\ I = \int \frac{2sin\left ( \frac{x}{2} \right )cos\left ( \frac{x}{2} \right )}{\sqrt{sin^2 \frac{x}{2}+cos^2 \frac{x}{2}}+ 2sin\left ( \frac{x}{2} \right )cos\left ( \frac{x}{2} \right )}dx \\\\\\ = 2\int\frac{sin\frac{x}{2}cos\frac{x}{2}}{|sin\frac{x}{2}cos\frac{x}{2}|} dx$

then depending on the values of 'x', he denominator will be + or -

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

HeroicPandas said:
$$Let $I = \int \frac{sinx}{\sqrt{1+sinx}}dx \\ \\ I = \int \frac{2sin\left ( \frac{x}{2} \right )cos\left ( \frac{x}{2} \right )}{\sqrt{sin^2 \frac{x}{2}+cos^2 \frac{x}{2}}+ 2sin\left ( \frac{x}{2} \right )cos\left ( \frac{x}{2} \right )}dx \\\\\\ = 2\int\frac{sin\frac{x}{2}cos\frac{x}{2}}{|sin\frac{x}{2}cos\frac{x}{2}|} dx$

then depending on the values of 'x', he denominator will be + or -

Wait doesn't it become:

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

?

rumbleroar · Jan 27, 2014

Re: MX2 Integration Marathon

Does anyone have relatively easier integration ones for MX2 noobs like me haha

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

rumbleroar said:
Does anyone have relatively easier integration ones for MX2 noobs like me haha

I'll make one on the spot. Have you done t-method yet?

rumbleroar · Jan 27, 2014

Re: MX2 Integration Marathon

anomalousdecay said:
I'll make one on the spot. Have you done t-method yet?

Yes (although I'm like really rusty)

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

rumbleroar said:
Yes (although I'm like really rusty)

I'll give something a bit simple then first up.

$\int^{\frac{\pi}{2}}_0 \frac{d\theta}{1+cos\theta + sin\theta}$

dunjaaa · Jan 27, 2014

Re: MX2 Integration Marathon

ln(2)

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

dunjaaa said:
ln(2)

Correct. Have a go at the one Sy posted before that I tried three times but didn't use the correct method for on the previous page.

I'm not sure if RR knows what to do (just ask for help).

dunjaaa · Jan 27, 2014

Re: MX2 Integration Marathon

Similar idea as the previous one

dunjaaa · Jan 27, 2014

Re: MX2 Integration Marathon

maybe she hasnt learnt inverse trig yet

anomalousdecay · Jan 27, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 29858 Similar idea as the previous one

Now I know where I failed.

I forgot that:

$cot(\frac{\pi}{2} - x) = tan(x)$

Oh well that's what happens 3 months after HSC haha.

Nice work

dunjaaa said:
maybe she hasnt learnt inverse trig yet

You only need t-method. No inverse trig is required.

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