P3 , integration problem, looks right but...

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Flakker
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#1
Report Thread starter 16 years ago
#1
Im integrating 27Sin(2x)Sin(x) between 0 and Pi/2.

I get the following:

int[27sin(2x)sin(x)]=int[27sin^2(x)cos(x)]=9sin^3(x)

as sin 0 =0 and sin pi/2 =1, my solution is 9 square units. however, an answer sheet I found gets the answer as 18 square units! can anyone shed any light on my error? Its probably my constant that is incorrect. btw, this is question 8. (c) of edexcel P3 June 2001.
The real thing on tues .

Thanks
Chris
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Rich
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#2
Report 16 years ago
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Originally posted by Flakker
Im integrating 27Sin(2x)Sin(x) between 0 and Pi/2.

I get the following:

int[27sin(2x)sin(x)]=int[27sin^2(x)cos(x)]=9sin^3(x)

as sin 0 =0 and sin pi/2 =1, my solution is 9 square units. however, an answer sheet I found gets the answer as 18 square units! can anyone shed any light on my error? Its probably my constant that is incorrect. btw, this is question 8. (c) of edexcel P3 June 2001.
The real thing on tues .

Thanks
Chris
27*Int(sin(x)sin(2x) dx) = 27*Int(2*sin^2(x)cos(x) dx)
= 54*Int(sin^2(x)cos(x) dx)
= 54*(1/3)*sin^3(x) + C
= 18*sin^3(x) + C

Putting in the limits:

Answer = 18*(sin^3(pi/2) - sin^3(0)) = 18*(1 - 0) = 18

It appears you left out a factor of two when using the trig. identity sin(2x) = 2sin(x)cos(x) (you incorrectly used sin(2x) = sin(x)cos(x)).

Good luck for Tuesday,

Regards,
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Flakker
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#3
Report Thread starter 16 years ago
#3
Thanks a lot!

/me smacks forehead.

Can't believe I lost that 2!

Chris
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