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Re: HSC 2016 4U Marathon $ a) After rigourous expansion and simplification the final integral is : $ 2 \int_0^{\pi}{\cos{(x(n+1))}}dx = 0 $ b) $ I_0 =0 , I_1=\pi $ c) Statement is true for n=0,1 (since) $ $ Assume n=k-1 and n=k is true $ I_{(k-1)}=\pi (k-1) , I_k= \pi k $...

There's only one way to find out... Look at HSC circle geometry questions.

Re: MX2 2016 Integration Marathon do you mean \frac{\log{(x+1)}}{x^2+1} ? Also the integral according to wolfram isn't the same as yours. EDIT: NVM they used ln not log my bad.

When are they open can I go now ?

Okay so what I did was enrolled into higher maths but without realising I didn't enroll in any tutes D: I can't change this (and I also want to change my lecture time.) I tried dropping it and it didn't let, I tried swapping it and it didn't let... It writes "Unable to swap course MATH1141...

Re: MX2 2016 Integration Marathon Yeah I didnt see this because his original integral was indefinite.

Re: MX2 2016 Integration Marathon Hint because I'm short on time: $ (1) Use IBP originally with $ u=x $ (2) Integrate v by letting $ \tan{\theta} =x $ (3) This can now be integrated by the use of two IBP's $ $ (4) Now to integrate $ -u'v $ We also do integration by parts twice $...

Re: MX2 2016 Integration Marathon $ Here's a little easy one I thought of just now : $ \int{(tan^{-1}x)^n}dx

Re: MX2 2016 Integration Marathon We pray 5 times a day :P And I havn't sleept yet

Re: MX2 2016 Integration Marathon The reduction formula for these integrals appear the same is there a general rule to memorise these or are they only done through rigorous expansion and +- techniques?

Re: MX2 2016 Integration Marathon LOL he knows uni stuff... Dw the only way you can beat him is by him making silly mistakes:P Since HSC is easy its possible that both of you could get 100:P

Re: HSC 2016 MX2 Combinatorics Marathon Doesn't matter this is clearly a 3U question level.

Re: HSC 2016 MX2 Combinatorics Marathon I think this thread is an extension 2 combinations thread.

Re: HSC 2016 2U Marathon $ Hence or otherwise find: $ \int{\cos{(2x)}{(\pi ^{\sin ^2 x}+ e ^{\sin ^2 x})^2}}dx $ Given that $ \int{\pi ^{2u} (1+e^{2u})}du $ can be expressed as $ \int{(\pi ^{2 \sin ^2x}+ \pi ^{2 \sin ^2 x} e^{2 \sin ^2 x}) \cos{2x}}dx $ where $ u= \sin ^2 x

Re: HSC 2016 4U Marathon n/2 is correct

What's wrong with calling bad reactions and consequences punishments and good reactions and consequences rewards? It's just the way Ustadh Said Nursi says it so I say it that way too :P

Exactly the reward for an antidote is good health I don't know how the laws of the Universe is that hard to understand :P ... The cure would be prevention of contact as in not getting close to it in the first place.

Re: HSC 2016 3U Marathon There's no use of it in any HSC level. (except maybe general notsure about the syllabus.)

That's me observing everything in a positive way :P What do you mean you don't agree ?... You don't agree that drinking poison is harmful :P ?

Re: HSC 2016 3U Marathon Here's a question before I forget : $ Solve: $ y=x^3+x^2-4x+3---(1) (y+1)(y^2-4)=x-7 ---(2)