Author Topic: 4U Maths Question Thread (Read 665231 times) Tweet Share

massive · « **Reply #405 on:** August 15, 2016, 10:49:09 am »

sweeeet thanks Rui!!

Also can i just ask another question. How do you do part ii for the q attached. Obviously you have to use part i because 'hence' but how? What i was thinking was to use by parts but I don't think that's gonna work.

RuiAce · « **Reply #406 on:** August 15, 2016, 10:57:06 am »

Quote from: massive on August 15, 2016, 10:49:09 am

sweeeet thanks Rui!!

Also can i just ask another question. How do you do part ii for the q attached. Obviously you have to use part i because 'hence' but how? What i was thinking was to use by parts but I don't think that's gonna work.

I think I've seen this question before. It's been tagged as faulty.

Yeah, IBP isn't going to work because to integrate the sqrt(x^2-4) we need an x in the numerator. But by introducing that x we just made the question substantially more messy.

massive · « **Reply #407 on:** August 15, 2016, 11:25:12 am »

Quote from: RuiAce on August 15, 2016, 10:57:06 am

It's been tagged as faulty.

So the questions wrong?

RuiAce · « **Reply #408 on:** August 15, 2016, 01:28:47 pm »

Quote from: massive on August 15, 2016, 11:25:12 am

So the questions wrong?

Yeah. Pretty sure it should ∫xLn(x²-4)/√(x²-4)dx

massive · « **Reply #409 on:** August 15, 2016, 02:35:23 pm »

how do you do this man ?

jakesilove · « **Reply #410 on:** August 15, 2016, 03:03:16 pm »

Quote from: massive on August 15, 2016, 02:35:23 pm

how do you do this man ?

Hey!

My solution is below. A bit of a trekk for an integration question, but really it was just brute force. I hope my working makes sense!

Jake

massive · « **Reply #411 on:** August 15, 2016, 04:03:35 pm »

Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction I_n ?

jakesilove · « **Reply #412 on:** August 15, 2016, 05:35:11 pm »

Quote from: massive on August 15, 2016, 04:03:35 pm

Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction I_n ?

Hey,

I'm on the run so won't be able to write up a full solution right now, but I think you have to break up the integral into the thing in part i), and then whatever is left over (I think x^(n-1)). Then, integrate using i), differentiate, preso the answer should pop out!

Jake

RuiAce · « **Reply #413 on:** August 15, 2016, 06:24:21 pm »

Quote from: massive on August 15, 2016, 04:03:35 pm

Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction I_n ?

$\text{Use part a) - You already have a primitive for }x\tan^{-1}x\\ \text{Integrate the whole of that in your IBP.}\\ \text{Also, try to break things into separate sections. This is quite an obliterating reduction formula.}$
$\begin{align*}I_n&=\int_0^1x^n\tan^{-1}x\,dx\\&=\int_0^1(x^{n-1})(x\tan^{-1}x)dx\\ &= \left[x^{n-1}\left(\frac{1}{2}(x^2+1)\tan^{-1}x-\frac{1}{2}x\right)\right]_0^1 - \int_0^1 nx^{n-2}\left(\frac{1}{2}(x^2+1)\tan^{-1}x-\frac{1}{2}x\right) dx\\ &= 2\tan^{-1}1-\frac{1}{2}-n\int_0^1\left(\frac{1}{2}(x^{n}+x^{n-2})\tan^{-1}x-\frac{1}{2}x^{n-1}\right)dx\\ &= \frac{\pi}{2}-\frac{1}{2}-\frac{n}{2}\int_0^1 (x^{n}+x^{n-2})\tan^{-1}x\, dx+\frac{n}{2}\int_0^1x^{n-1}\, dx\end{align*}$
$\text{I'll let you take over here.}$

massive · « **Reply #414 on:** August 15, 2016, 06:47:18 pm »

Thanks Rui! yeah i eventually got it

Guys check out the integration q attached. Its pretty interesting never seen anything like it. how does one go about solving something like that. At first I thought partial fractions but i don't think that's going to work. Im guessing substitution, idek. Any thought??

RuiAce · « **Reply #415 on:** August 15, 2016, 08:14:28 pm »

Quote from: massive on August 15, 2016, 06:47:18 pm

Thanks Rui! yeah i eventually got it

Guys check out the integration q attached. Its pretty interesting never seen anything like it. how does one go about solving something like that. At first I thought partial fractions but i don't think that's going to work. Im guessing substitution, idek. Any thought??

I briefly looked at it and first deduced that it had gone well beyond the scope of the HSC as any exponentiation on itself is tremendously hard to handle. I.e. x^x

But after spending 4 more minutes on it I realised wait no it's too ridiculous

massive · « **Reply #416 on:** August 15, 2016, 08:18:47 pm »

Quote from: RuiAce on August 15, 2016, 08:14:28 pm

(Image removed from quote.)

I briefly looked at it and first deduced that it had gone well beyond the scope of the HSC as any exponentiation on itself is tremendously hard to handle. I.e. x^x

But after spending 4 more minutes on it I realised wait no it's too ridiculous

Thanks mate!!

massive · « **Reply #417 on:** August 15, 2016, 08:28:02 pm »

how do you do part ii guys?

RuiAce · « **Reply #418 on:** August 15, 2016, 08:29:52 pm »

Quote from: massive on August 15, 2016, 08:28:02 pm

how do you do part ii guys?

Hint (since I'm having dinner right now)

Let x=tan(theta)

Inspiration: because the derivative of tan is sec squared and because 1+tan^2=sec^2 AND the boundaries were too convenient

massive · « **Reply #419 on:** August 15, 2016, 08:42:45 pm »

Quote from: RuiAce on August 15, 2016, 08:29:52 pm

Hint (since I'm having dinner right now)

Let x=tan(theta)

Inspiration: because the derivative of tan is sec squared and because 1+tan^2=sec^2 AND the boundaries were too convenient

But then how do you get a reduction formula using that :S

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665231 times) Tweet Share

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

4U Maths Question Thread

massive

Re: 4U Maths Question Thread

Recent Posts

User:
Password: