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# P3 maths integration question watch

1. hey guys. i was wondering is all the good mathematicians out there could help me on a hard question. (if it helsp this question is question 7 on an OCR past P3 paper dated 16 January 2002). i dont have a clue on thsi so could someone go through it step by sytep.

by using substitution y=a sinx to prove that the

\$ (limits a and a/2) (a² - y²)^½ dx = (a²/2)[(n/3) - (3^½)/2]

\$ means integral
^½ means to the power of a half
n means pi (i.e. 3.14)
2. you've written dx, don't yoou mean dy?

y = a sinx => dy/dx = a cosx

\$ (limits a and a/2) (a² - y²)^½ dy = \$ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= \$ (limits pi/6 and pi/2) a² cos²x dx
= \$ (limits pi/6 and pi/2) a²/2 (1+cos2x) dx

=a²/2[x + 1/2 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)
3. What does the dollar sign mean?
4. yeah sorry i meant dy not dx.

by the way the dollar sign means intergral. u know... the squigly line thing !
5. (Original post by elpaw)
you've written dx, don't yoou mean dy?

y = a sinx => dy/dx = a cosx

\$ (limits a and a/2) (a² - y²)^½ dy = \$ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= \$ (limits pi/6 and pi/2) a² cos²x dx
= \$ (limits pi/6 and pi/2) a/2 (1+cos2x) dx

=[ax/2 + a/4 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)
cheers elpaw that was great! u did it just like that! iwas expecting a few hours before someone got the answer for me. no wonder ur at oxford!. i bow down to ur greatness.
6. however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
7. (Original post by DLo)
however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
The original limits are wrt y. The new limits are wrt x.

using the substitution y=asinx,

at y=a/2, a/2=asinx => sinx = ½ => x=pi/6
at y=a, a=asinx => sinx = 1 => x=pi/2

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