Does anyone can proof the ∫ cos x dx = sin x + C?

karakoos23 · Nov 26, 2006

Hi

Does anyone can proof the ∫ cos x dx = sin x + C?

Thanks in advance

bunalmis · Nov 26, 2006

Re: Integration problem

2Cos(x)=e^jx +e^-jx

2jSin(x)=e^jx -e^-jx

∫ cos x dx = 1/2 ∫ e^jx dx +1/2 ∫ e^-jx dx = =1/2j e^jx - 1/2j e^-jx + C = Sin(x) + C

Added after 1 hours 52 minutes:

∫ cos x dx = Sin(x) + C

d/dx (Sin(x) + C) = ?

d/dx Sin(x) = lim dx-->0 (1/dx) (Sin(x+dx) - Sin(x)) = lim dx-->0 (Sin(x)Cos(dx) +Sin(dx)Cos(x) - Sin(x))/dx

lim dx-->0 Sin(x)Cos(dx)/dx - Sin(x)/dx = 0

lim dx-->0 Sin(dx)Cos(x) /dx = cos(x)

d/dx Sin(x)=Cos(x)

karakoos23 · Nov 27, 2006

Re: Integration problem

Hi bunalmis

Can you please tell me what title of textbook in which I can find that analysis of cosinus?

It would be much appreciated

free1965tv · Mar 17, 2007

Re: Integration problem

Wow

Added after 39 seconds:

bunalmis said:
2Cos(x)=e^jx +e^-jx

2jSin(x)=e^jx -e^-jx

∫ cos x dx = 1/2 ∫ e^jx dx +1/2 ∫ e^-jx dx = =1/2j e^jx - 1/2j e^-jx + C = Sin(x) + C

Added after 1 hours 52 minutes:

∫ cos x dx = Sin(x) + C

d/dx (Sin(x) + C) = ?

d/dx Sin(x) = lim dx-->0 (1/dx) (Sin(x+dx) - Sin(x)) = lim dx-->0 (Sin(x)Cos(dx) +Sin(dx)Cos(x) - Sin(x))/dx

lim dx-->0 Sin(x)Cos(dx)/dx - Sin(x)/dx = 0

lim dx-->0 Sin(dx)Cos(x) /dx = cos(x)

d/dx Sin(x)=Cos(x)

Wow

Does anyone can proof the ∫ cos x dx = sin x + C?

karakoos23

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bunalmis

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karakoos23

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free1965tv

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