Competition - post a photo you've taken of nature >>

STEP 1994 Question

So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

It's q 4ii) on 1994 STEP I (attached below)

I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

Am I just not using the method correctly?

:frown:

Reply 1

Blutooth

Original post by shanban

So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

It's q 4ii) on 1994 STEP I (attached below)

I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

Am I just not using the method correctly?

:frown:

You need to use the sub cosalpha =k to get rid of the sinalpha at the top.

Eidt: forget what I said was talking nonsense.

(edited 11 years ago)

Reply 2

shanban

Original post by Blutooth

You need to use the sub cosalpha =k to get rid of the sinalpha at the top.

Eidt: forget what I said was talking nonsense.

I was just wondering why it wouldn't work without a trig sub...but thanks anyway! :smile:

Reply 3

M1K3

Original post by shanban

So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

It's q 4ii) on 1994 STEP I (attached below)

I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

Am I just not using the method correctly?

:frown:

Are we on about the first part of ii)?

We complete the square to get

\displaystyle\int \dfrac{1}{(x-k)^2 + (1-k^2)}\ dx

are u alrite with this or is it something else?

Reply 4

shanban

Original post by M1K3

Are we on about the first part of ii)?

We complete the square to get

\displaystyle\int \dfrac{1}{(x-k)^2 + (1-k^2)}\ dx

are u alrite with this or is it something else?

Yeah I got that, and then I tried to make it into an integral which would then give something along the lines of arctan...but the constant I took out of the integral didn't have a square root where it should :frown:

Reply 5

TheJ0ker

Original post by shanban

Have you tried a half tan substitution? I think that is what the question is looking for.

EDIT: in fact if you have

\displaystyle\int \frac{1}{(x-k)^2 + (\sqrt{1-k^2})^2} dx

then can you see what to do?

(edited 11 years ago)

Reply 6

shanban

Original post by TheJ0ker

Have you tried a half tan substitution? I think that is what the question is looking for.

EDIT: in fact if you have

\displaystyle\int \frac{1}{(x-k)^2 + (\sqrt{1-k^2})^2} dx

then can you see what to do?

This is what I've done:

\dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx [br][br]= \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C[br]

I don't understand why you'd put the sqrt in?

(edited 11 years ago)

Reply 7

TheJ0ker

Original post by shanban

This is what I've done:

\dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx [br][br]= \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C[br]

I don't understand why you'd put the sqrt in?

Woah why are you taking stuff out of the integral? Do you know what the standard result is? It's on the formula book FP3 part.

Reply 8

M1K3

Original post by shanban

This is what I've done:

\dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx [br][br]= \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C[br]

I don't understand why you'd put the sqrt in?

As theJoker said, this is a standard integral.

But your way is fine, the reason why your method has not worked is because i think you mite have done the integration slightly wrong, well, just differentiate your answer and i think you'll understand.

Remember, you need to divide by the differential of