# P3...Integration.....

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#1
Integrate sin3xcos2x dx

0
17 years ago
#2
I would suggest using parts. I suspect you'd have to integrate by parts again, and still remain with an integral, but that integral would be similar to the question...
0
17 years ago
#3
(Original post by imasillynarb)
Integrate sin3xcos2x dx

apply:
2sinAcosB≡sin(A+B)+sin(A-B) to sin3xcos2x
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#4
(Original post by IntegralAnomaly)
apply:
2sinAcosB≡sin(A+B)+sin(A-B) to sin3xcos2x
How do you KNOW that this identity will work and others wont? This is really pissing me off now
0
17 years ago
#5
(Original post by imasillynarb)
How do you KNOW that this identity will work and others wont? This is really pissing me off now
????? i dont understand ur query, i logically this identity should be ur first thought:
sin3xcos2x=1/2(sin5x+sinx)
so ∫sin3xcos2x dx = 1/2(-1/5cos5x-cosx)
0
17 years ago
#6
(Original post by imasillynarb)
How do you KNOW that this identity will work and others wont? This is really pissing me off now
by the way remain patient with maths,espiecally on integration.Dont get angry cos u wont be able to do the problem then
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#7
(Original post by IntegralAnomaly)
????? i dont understand ur query, i logically this identity should be ur first thought:
sin3xcos2x=1/2(sin5x+sinx)
so ∫sin3xcos2x dx = 1/2(-1/5cos5x-cosx)

OK, so, if the question is, cos2xcos5x - why wouldnt I use the identity:

2cosAcosB = cos(a+B) +cos(A-B) ??
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17 years ago
#8
(Original post by imasillynarb)
OK, so, if the question is, cos2xcos5x - why wouldnt I use the identity:

2cosAcosB = cos(a+B) +cos(A-B) ??
well you would use that:
∫cos2xcos5xdx=1/2∫cos7xdx+1/2∫cos(-3x)dx (cos(-3x) is of course ≡ cos3x)
so ∫cos2xcos5xdx=(1/14)sin7x+(1/6)sin3x
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#9
(Original post by IntegralAnomaly)
well you would use that:
∫cos2xcos5xdx=1/2∫cos7xdx+1/2∫cos(-3x)dx (cos(-3x) is of course ≡ cos3x)
so ∫cos2xcos5xdx=(1/14)sin7x+(1/6)sin3x
Ahar, nice one.
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17 years ago
#10
Does anyone hav OCR P3 mark scheme?(june 2001)
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17 years ago
#11
the answer is (-4/5)*((1/2)*sin(3*x)*sin(2*x) + (3/4)*cos(3*x)*cos(2*x))
which can be simplified to the above whoever suggested it.
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17 years ago
#12
(Original post by cel)
Does anyone hav OCR P3 mark scheme?(june 2001)
This thread http://www.uk-learning.net/t25640.html has got all the links to the exam board's sites. Great idea... very useful!
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