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Re: HSC 2016 MX2 Combinatorics Marathon ftfy.

Almost. I am saying that a good action of equivalent value is considered greater if it done with conscience, free-will and good-intention. We believe that everyone has a "jinn" 'inside' them which 'whispers' to us to do bad things. So this being is always whispering bad things trying to...

Because creating something other than angels with free will as I said allows us to surpass the angels. Also God didn't need to create humans, as you probably asserted through the word "necessary." Also we believe that before entering the world God asked us if we wanted to be human so we cant...

Sorry ill try again :P So what we believe in(again it's a belief you have a right to not agree and I have convinced myself with proofs that I agree with that probably everyone else won't) is that if there is a choice/inclination to do bad then it would obviously be more valuable to not do that...

Hes already made a perfect creation; the angels if He did make us perfect there would be no difference between us and angels. It allows us to be greater than angels as I said or worse than the devils. :)

To feel bad about what you have done, make the intention to never do the sin again and try your best not to do it again.

One of the signs of the coming of judgement day is that most miracles will be able to be performed by people but not to the extent that the original miracles were. Well this is the Islamic belief you don't have to accept it :).

Here is the Muslim stance: There is good in everything God creates. He created everything with the ability to either do either good or bad. In regards to the Satan, unlike the Christians who believe him to be a "fallen angel" (please correct me if I am mistaken), we believe him to be a "Jin"...

Re: MATH1231/1241/1251 SOS Thread In b4 thread closes

Re: MX2 2016 Integration Marathon Well what I did was let : u = e^x - \frac{e^b + e^a}{2} $ Then $ u = (\frac{e^b+e^a}{2}) \sin{\theta} $ My integral was an easy $ \frac{1}{1 + \sin{\theta}}

Re: MX2 2016 Integration Marathon I got something like (assuming no mistakes i'm pretty sure my method was correct). : I = \frac{4(b-a)}{(e^b+e^a) \cdot \sqrt{(e^b+e^a)^2 -(b-a)^2}} + C

Re: HSC 2016 4U Marathon Yes.

Re: HSC 2016 4U Marathon Somewhat difficult if you don't know the 'guitar method.' $ Solve : $ |x| - |7-x| +2|x-2| \leq 4

Re: MX2 2016 Integration Marathon Since this isn't a very common type of questions ill include any thoughts I had. ( Excuse any mistakes pls). $ I noticed that the denominator had a square in it and thus may have something to do with the quotient rule. However the form is too complex to...

Re: MX2 2016 Integration Marathon $ Let x = \cos{\theta} I = \int{\tan^2{\theta} \frac{1}{1 - 2\cos^2{\theta}}}d\theta $ Dividing by $ \cos^2{x} I = \int{\frac{\tan^2{\theta}\sec^2{\theta}}{(\tan^2{ \theta } - 1)}}d\theta u = \tan{\theta} I = \int{1 +...

Re: MX2 2016 Integration Marathon $ Let $ x = \tan{(\theta)} \therefore I = \int_0^{\tan^{-1}a}{\log{(1+a\tan{\theta})}d\theta} $ Using $ \int_0^a{f(x)}dx = \int_0^a{f(a-x)}dx I = \int_0^{\tan^{-1}a}{\log{(\frac{1+a^2}{1+a\tan{\theta}})}d\theta} 2I =...

Re: MX2 2016 Integration Marathon $ Let $ x^t = u I = \frac{1}{t} \int{\frac{1+au}{u(a+bu^2)}}du $ By applying partial fractions we notice that $ I = \frac{1}{t}\int{(\frac{1}{u} + \frac{a}{1+bu^2} - \frac{bu}{1+bu^2})}du I = ln|x| -\frac{1}{2t} \ln{|1+bx^{2t}|} +...

Re: MX2 2016 Integration Marathon Yes as if you didn't notice the hint. It's called "YOLO."

Re: MX2 2016 Integration Marathon $ Not too sure about the other values of $ \alpha $ but will attempt soon. $ \int_0^{\pi}{\ln{(1- 2\alpha \cos{x} + \alpha ^ 2)}}dx $ Take note that $ |\alpha| = 1 $ means that $ \alpha ^2 = 1 I = \int_0^{\pi}{\ln{(2- 2\alpha \cos{x})}}dx $ Using...

Re: MX2 2016 Integration Marathon Excuse the mistakes if any. $ Multiply top and bottom by $ (\sin{x} + 1) - \cos{x}. I = \frac{1}{2} \int{\frac{(\sin{x}+1)^2 - \cos{x}(1+\sin{x})}{\sin{x} + 1}}dx I = \frac{1}{2} \int{\sin{x} - \cos{x} + 1}dx I = \frac{x}{2} - \frac{\sin{x} +...