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

dunjaaa · Sep 10, 2014

Re: MX2 Integration Marathon

so the answer should be e^4.6 which is 99.95 (2dp)

dan964 · Sep 10, 2014

Re: MX2 Integration Marathon

dunjaaa said:
so the answer should be e^4.6 which is 99.95 (2dp)

Nice. I see what you did there.

dan964 · Sep 10, 2014

Re: MX2 Integration Marathon

Integrate tan^2(x)*e^x

glittergal96 · Sep 16, 2014

Re: MX2 Integration Marathon

dan964 said:
No it is e to the power of x times e to the power of x^2

dan964 said:
Integrate tan^2(x)*e^x

I don't think either of these functions have a primitive that can be expressed with elementary functions?

Fade1233 · Sep 16, 2014

Re: MX2 Integration Marathon

dunjaaa said:
I remember doing this question in a trial paper . I'll post a solution tomorrow if nobody has solved it yet. Hint: Use a trig-substitution

Similar to BOS 2012 Paper question

dan964 · Sep 16, 2014

Re: MX2 Integration Marathon

Integrate 2 ln (1+x^2/1-x^2) with respect to x

dunjaaa · Sep 16, 2014

Re: MX2 Integration Marathon

Screen shot 2014-09-16 at 6.48.11 PM.png

, nothing too hard, just tedious.
Next:

Screen shot 2014-09-16 at 9.11.55 PM.png

integral95 · Sep 16, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30918, nothing too hard, just tedious.
Next:View attachment 30920

$$let I_{2n+1} = $ \int_{0}^{\frac{\pi}{2}} sin^{2n+1}x \\ \\ $using integration by parts with u = sin^{2n}x \ \ dv = sinx $ \\ \\ I_{2n+1} = 2n\int_{0}^{\frac{\pi}{2}}sin^{2n-1}x - sin^{2n+1}x \ dx \\ = 2n(I_{2n-1} - I_{2n+1}) \\ \\ \Rightarrow I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} = \\ \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)}I_1 \\ \\ I_1 = 1 \\\therefore I_{2n+1} = \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)} = \frac{2^n(n!)}{(2n+1)(2n-1)(2n-3)...(3)} \\ = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$

Not sure if i made a mistake there lol, and yeah i know i skipped a lot towards the end.

As for the second part I manipulated the first equation to make it look like the expression in the question and took the summation but i'm not sure what to do there.

Speed6 · Sep 16, 2014

Re: MX2 Integration Marathon

integral95 said:
$$let I_{2n+1} = $ \int_{0}^{\frac{\pi}{2}} sin^{2n+1}x \\ \\ $using integration by parts with u = sin^{2n}x \ \ dv = sinx $ \\ \\ I_{2n+1} = 2n\int_{0}^{\frac{\pi}{2}}sin^{2n-1}x - sin^{2n+1}x \ dx \\ = 2n(I_{2n-1} - I_{2n+1}) \\ \\ \Rightarrow I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} = \\ \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)}I_1 \\ \\ I_1 = 1 \\\therefore I_{2n+1} = \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)} = \frac{2^n(n!)}{(2n+1)(2n-1)(2n-3)...(3)} \\ = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$

Not sure if i made a mistake there lol, and yeah i know i skipped a lot towards the end.

As for the second part I manipulated the first equation to make it look like the expression in the question and took the summation but i'm not sure what to do there.

Your smart.

Fade1233 · Sep 16, 2014

Re: MX2 Integration Marathon

integral95 said:
$$let I_{2n+1} = $ \int_{0}^{\frac{\pi}{2}} sin^{2n+1}x \\ \\ $using integration by parts with u = sin^{2n}x \ \ dv = sinx $ \\ \\ I_{2n+1} = 2n\int_{0}^{\frac{\pi}{2}}sin^{2n-1}x - sin^{2n+1}x \ dx \\ = 2n(I_{2n-1} - I_{2n+1}) \\ \\ \Rightarrow I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} = \\ \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)}I_1 \\ \\ I_1 = 1 \\\therefore I_{2n+1} = \frac{(2n)(2n-2)(2n-4)...(2)}{(2n+1)(2n-1)(2n-3)...(3)} = \frac{2^n(n!)}{(2n+1)(2n-1)(2n-3)...(3)} \\ = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$

Not sure if i made a mistake there lol, and yeah i know i skipped a lot towards the end.

As for the second part I manipulated the first equation to make it look like the expression in the question and took the summation but i'm not sure what to do there.

(2^(2n+1))/(2n+1)(2nCn)= what you had proven.
Doesnt that work?

aDimitri · Sep 16, 2014

Re: MX2 Integration Marathon

Speed6 said:
Your smart.

you're*
( ͡° ͜ʖ ͡°)

integral95 · Sep 17, 2014

Re: MX2 Integration Marathon

Fade1233 said:
(2^(2n+1))/(2n+1)(2nCn)= what you had proven.
Doesnt that work?

$I_{2n+1} \frac{2n+1}{2^{2n+1}} = $that question$$

But after taking the sum likethe question states, I dont know how to simplify it... what am I missing haha

dan964 · Sep 17, 2014

Re: MX2 Integration Marathon

here is one from a friend of mine that tests all your MX2 Integration skills
integrate with respect to x
2 ln (1+x/1+x^2)

Carrotsticks · Sep 17, 2014

Re: MX2 Integration Marathon

dan964 said:
here is one from a friend of mine that tests all your MX2 Integration skills
integrate with respect to x
2 ln (1+x/1+x^2)

?

Just split the log and then parts for each term.

dan964 · Sep 17, 2014

Re: MX2 Integration Marathon

Carrotsticks said:
?

Just split the log and then parts for each term.

sure, it still tests all your skills even if it doesn't cover everything. I'd didn't say it was going to be hard.
(you have to be able to integrate by parts effectively twice, and trigonometry)

Carrotsticks · Sep 17, 2014

Re: MX2 Integration Marathon

dan964 said:
sure, it still tests all your skills even if it doesn't cover everything. I'd didn't say it was going to be hard.
(you have to be able to integrate by parts effectively twice, and trigonometry)

From observation (I haven't actually done it yet), the only MX2 technique required is IBP, which is hardly 'all your skills' considering there are perhaps 5-6 other integration techniques in Extension 2.

integral95 · Sep 17, 2014

Re: MX2 Integration Marathon

right......................

$\int_{0}^{\frac{\pi}{4}} \sum\limits_{k=0}^\infty sin^{2k+1}x dx$

dunjaaa · Sep 18, 2014

Re: MX2 Integration Marathon

yeah there should also be a (2n+1)/2^(2n+1) outside and expand the sigma

integral95 · Sep 18, 2014

Re: MX2 Integration Marathon

dunjaaa said:
yeah there should also be a (2n+1)/2^(2n+1) outside and expand the sigma

I don't think yiu can take it outside of sigma since the expression is in terms of n

Sent from my C5303 using Tapatalk

dunjaaa · Sep 18, 2014

Re: MX2 Integration Marathon

in your case, it'll be in terms of 'k' [i.e. (2k+1)/2^(2k+1)], since that's what you chose to represent as the index of summation

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