4U Maths Question Thread

[bluecookie]

thanks

[bluecookie]

If In=integral tan^nx dx for n>=0 show that In=[1/(n-1)]*tan^(n-1)x-I(n-2) for n>=2. What's the setting out for proving n>=2.

[RuiAce]

If In=integral tan^nx dx for n>=0 show that In=[1/(n-1)]*tan^(n-1)x-I(n-2) for n>=2. What's the setting out for proving n>=2.

The normal setting out.

Note that for the integral of tan^{n[/sup(x) you should not be using by parts, but only a Pythagorean identity.}

[bluecookie]

The normal setting out.

Note that for the integral of tan^{n[/sup(x) you should not be using by parts, but only a Pythagorean identity.}

Thanks!

Show that for In=integral sec^nxdx >=0, In=1/(n-1)tanxsec^n-2x+(n-2)/(n-1)*I(n-2) for n>=2.

For these type questions, how do we know which way to start? Because I can see several ways this question might work. (take out sec squared which gives sec^n-2 which matches the I(n-2) expression a little, or change it into secx*sec^n-1x and differentiate the sec^n-1x to get the I(n-2) (I tried this but got a really long expression that I was stuck unsure of what to do with) etc) but how do I know is the right method?

[RuiAce]

Thanks!

Show that for In=integral sec^nxdx >=0, In=1/(n-1)tanxsec^n-2x+(n-2)/(n-1)*I(n-2) for n>=2.

For these type questions, how do we know which way to start? Because I can see several ways this question might work. (take out sec squared which gives sec^n-2 which matches the I(n-2) expression a little, or change it into secx*sec^n-1x and differentiate the sec^n-1x to get the I(n-2) (I tried this but got a really long expression that I was stuck unsure of what to do with) etc) but how do I know is the right method?

These really standard ones are ones you should memorise the method of.

For that one, you should consider \(\sec^2x\) and \(\sec^{n-2}x\) because of the relevant Pythagorean identity. But you will need integration by parts - Only questions that involve \(\tan^n x\) or can be made into that form by a u-substitution can fully avoid IBP as far as I know in 4U.

[bluecookie]

Thanks!

[limtou]

Hello, please help with d) iv)
Thanks

[RuiAce]

Hello, please help with d) iv)
Thanks

I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.


One possible plan for the sketch:

1. Sketch \(x^2-y^2=3\) and \(xy = 2\) separately
2. Identify the regions satisfying \(x^2-y^2<3\) and \(xy<2\)
3. Add in the restriction \( 0 < x^2-y^2\) and \(0 < xy\) using your knowledge of the 4 quadrants. You may wish to do this separately from the rough sketches you had in step 2.
4. Combine the regions required.

[bluecookie]

If In=integral from 0 to 1 of x(1-x^3)^n for n>=0 show that In=[3n/(3n+2)]*I(n-1) for n>=1. Hence find an expression for In in terms of n for n>=0.

[RuiAce]

If In=integral from 0 to 1 of x(1-x^3)^n for n>=0 show that In=[3n/(3n+2)]*I(n-1) for n>=1. Hence find an expression for In in terms of n for n>=0.

[BradMate]

Could someone please explain how to solve this polynomial question. Thanks so much.

[limtou]

I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.


One possible plan for the sketch:

1. Sketch \(x^2-y^2=3\) and \(xy = 2\) separately
2. Identify the regions satisfying \(x^2-y^2<3\) and \(xy<2\)
3. Add in the restriction \( 0 < x^2-y^2\) and \(0 < xy\) using your knowledge of the 4 quadrants. You may wish to do this separately from the rough sketches you had in step 2.
4. Combine the regions required.

Thank you! I realised that I wasn't able to get the answer at first because I didn't consider the quadrants, but now I got it

[RuiAce]

Could someone please explain how to solve this polynomial question. Thanks so much.

No LaTeX because I'm at the gym, just a description: Edit; LaTeX in place now.

As the coefficients are integers, if a surd is a root then so must it's irrational conjugate. Here, that would be \(-\sqrt3\).

Then, as the polynomial is monic it must have leading coefficient is 1. By the product of roots, if the third root is \(\alpha\) we must have
\(\sqrt{3}\times -\sqrt3\times \alpha=-12 \implies \alpha=4\)

So the polynomial is
\(P(x)=(x-\sqrt3)(x+\sqrt3)(x-4)=\left(x^2-3\right)(x-4)\)

You may expand this out if you wish.

Thank you! I realised that I wasn't able to get the answer at first because I didn't consider the quadrants, but now I got it

Ah yep, that was most likely the hardest bit about that question.

[Ali_Abbas]

Could someone please explain how to solve this polynomial question. Thanks so much.

Here is my solution.

[RuiAce]

Here is my solution.

Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...