Quick Maths Problem

Scoobydooo

Ran into this question when going through the papers its on integration

evaluate: square root of 4 - x2 letting x = 2SINQ


Limits are 0 and root 3

the 2 in x2 is meant to be the squared sign and Q stands for theta

Thanks in advance !!

andyman

It's one of the earlier years isn't it?

Circular integration that sounds like. Can't quite remember how to do them but you have to let u = 2sinθ then differentiate, I think.

Could be wrong. If you use Aidan Roantrees book there's a section on it in the Integration chapter

.:FuZion:.

Im trying to do it. Stuck as far as (4cos(Squared)Q)^1/2(2cosQ)dQ.... :rolleyes:

Scoobydooo

Ye i think its from the 1999 papers any help is greatly appreciated

Des23

Fill in 2sinQ into x^2 and you also have to fill it into dx, so you have to differentiate to get 2cosQdQ (change the limits) after that just do all the trig and you have to integrate 2(1 + cos2Q) between pi/3 and 0

.:FuZion:.

Forgot to change the limits. :rolleyes:

Challenged

http://studentxpress.ie/papers/intsoln1999.pdf

Scoobydooo

Thanks for all your help guys !!

.:FuZion:.

Did you get it out? :pac: Im doing 2004 now.