Last question on AQA Core 3

You're right but I think you need to convert it back to algebra form in the previous part of the question. I got the same answer but I realised after the exam that when you had to prove that sqrt(1-4x²)/2x=cot(u). You have to flip it.

Reply 2

Hopefully we'll get the marks because we're not technically wrong and we did what was asked

Original post by Connor09

Find integral of 1/(1-4X^2)^(3/2), using the substitution X=1/2sin(u).

Just got the question in my exam, my answer was 1/2tan(Sin^(-1)(2X)). Which just seems a little odd, anybody know whether it’s right or not?

Picture should make it clearer.

Thanks
4800F904-E4CD-4639-9146-7AC5D82E22C0.jpg.jpeg

Reply 3

From what I remember, I think the answer would be X/sqrt(1-4x²)

Original post by Connor09

Find integral of 1/(1-4X^2)^(3/2), using the substitution X=1/2sin(u).

Just got the question in my exam, my answer was 1/2tan(Sin^(-1)(2X)). Which just seems a little odd, anybody know whether it’s right or not?

Picture should make it clearer.

Thanks
4800F904-E4CD-4639-9146-7AC5D82E22C0.jpg.jpeg

Reply 4

OP

Ahh yeah, I kept looking at the previous part but couldn’t see the connection properly. As you said the answer was in terms or X so hopefully we’ll get all the marks, if not only 1 or 2 lost.

Reply 5

Did you get like pi • [-(e^-1)-something(-e^-.5)-28) for the volume one?

Reply 6

OP

Haha, glad someone else did, that was another answer I thought looked funny. Can’t remember the solution exactly but it looked very similar, a 28 term and two e terms raised to powers of -1 and -0.5 and of course the pi. So hopefully we got that right

Reply 7

OP

Original post by Madethistohelp99

Did you get like pi • [-(e^-1)-something(-e^-.5)-28) for the volume one?

Haha, glad someone else did, that was another answer I thought looked funny. Can’t remember the solution exactly but it looked very similar, a 28 term and two e terms raised to powers of -1 and -0.5 and of course the pi. So hopefully we got that right

Reply 8

AldiiblA

9

does anyone know what the mark points will be for the 6 marker revolution q?

Original post by Connor09

Haha, glad someone else did, that was another answer I thought looked funny. Can’t remember the solution exactly but it looked very similar, a 28 term and two e terms raised to powers of -1 and -0.5 and of course the pi. So hopefully we got that right

Also the mark points for the integral question??

Original post by Connor09

Find integral of 1/(1-4X^2)^(3/2), using the substitution X=1/2sin(u).

Just got the question in my exam, my answer was 1/2tan(Sin^(-1)(2X)). Which just seems a little odd, anybody know whether it’s right or not?

Picture should make it clearer.

Thanks
4800F904-E4CD-4639-9146-7AC5D82E22C0.jpg.jpeg

Reply 9

OP