HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)
- Thread starter bleakarcher
- Start date
- Status
Carrotsticks
Retired
- Joined
- Jun 29, 2009
- Messages
- 9,494
- Gender
- Undisclosed
- HSC
- N/A
Re: MX2 Integration Marathon
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.
='(
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.
asianese
Σ
- Joined
- Sep 20, 2010
- Messages
- 2,225
- Gender
- Undisclosed
- HSC
- 2012
Re: MX2 Integration Marathon
I remember this from 2012 hahha. Leme see if I remember how to do it... (ceebs actually)This is always a good one
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
 \\ \\ \\ 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:
}{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:
 \ 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 )
(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.
This is always a good one
Well at least I know all my working out is correct. As for the method, not quite.
Second attempt:
Something is wrong with the working out here. I probably need to substitute instead. I'll give it one more go:
Its too messy for me now on latex. Also, the first part is completely incorrect as it gives me a tendency to negative infinity.
Last edited:
HeroicPandas
Heroic!
- Joined
- Mar 8, 2012
- Messages
- 1,547
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
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 -
then depending on the values of 'x', he denominator will be + or -
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
cos(\frac{x}{2})}{|sin( \frac {x}{2}) + cos (\frac{x}{2})|} $ $ dx )
?
Wait doesn't it become:
then depending on the values of 'x', he denominator will be + or -
?
rumbleroar
Survivor of the HSC
- Joined
- Nov 30, 2011
- Messages
- 2,271
- Gender
- Female
- HSC
- 2014
Re: MX2 Integration Marathon
Does anyone have relatively easier integration ones for MX2 noobs like me haha
Does anyone have relatively easier integration ones for MX2 noobs like me haha
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
I'll make one on the spot. Have you done t-method yet?
rumbleroar
Survivor of the HSC
- Joined
- Nov 30, 2011
- Messages
- 2,271
- Gender
- Female
- HSC
- 2014
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
I'm not sure if RR knows what to do (just ask for help).
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).
anomalousdecay
Premium Member
- Joined
- Jan 26, 2013
- Messages
- 5,766
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
I forgot that:
 = tan(x) )
Oh well that's what happens 3 months after HSC haha.
Nice work
Now I know where I failed.View attachment 29858 Similar idea as the previous one
I forgot that:
Oh well that's what happens 3 months after HSC haha.
Nice work
You only need t-method. No inverse trig is required.
- Status