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HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)
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Carrotsticks
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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
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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
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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
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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
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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
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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
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Re: MX2 Integration Marathon
I'll make one on the spot. Have you done t-method yet?
rumbleroar
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anomalousdecay
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anomalousdecay
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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
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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.
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