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P3...Integration.....
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#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...
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#3
(Original post by imasillynarb)
Integrate sin3xcos2x dx
Help please.
Integrate sin3xcos2x dx
Help please.
2sinAcosB≡sin(A+B)+sin(A-B) to sin3xcos2x
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(Original post by IntegralAnomaly)
apply:
2sinAcosB≡sin(A+B)+sin(A-B) to sin3xcos2x
apply:
2sinAcosB≡sin(A+B)+sin(A-B) to sin3xcos2x
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#5
(Original post by imasillynarb)
How do you KNOW that this identity will work and others wont? This is really pissing me off now
How do you KNOW that this identity will work and others wont? This is really pissing me off now
sin3xcos2x=1/2(sin5x+sinx)
so ∫sin3xcos2x dx = 1/2(-1/5cos5x-cosx)
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#6
(Original post by imasillynarb)
How do you KNOW that this identity will work and others wont? This is really pissing me off now
How do you KNOW that this identity will work and others wont? This is really pissing me off now
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(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)
????? 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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#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) ??
OK, so, if the question is, cos2xcos5x - why wouldnt I use the identity:
2cosAcosB = cos(a+B) +cos(A-B) ??
∫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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(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
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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#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.
which can be simplified to the above whoever suggested it.
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#12
(Original post by cel)
Does anyone hav OCR P3 mark scheme?(june 2001)
Does anyone hav OCR P3 mark scheme?(june 2001)
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