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Basic Integral without Basic Concept

  1. Nov 22, 2011 #1
    A friend proposed this problem to me.

    Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.
  2. jcsd
  3. Nov 22, 2011 #2


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    How about using the Taylor series expansion of the cosine function?[tex] \int \cos x\,dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\,dx[/tex][tex] = \sum_{n=0}^{\infty} \int \frac{(-1)^n x^{2n}}{(2n)!}\,dx [/tex][tex] = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\frac{1}{2n+1}x^{2n+1}[/tex]Now, (2n+1)*(2n)! = (2n+1)! so that this becomes[tex]\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}[/tex]Now all you have to do is "notice" that this result is the Taylor series expansion of a sine function.

    I just realized that you'd probably have to know the derivative of cosine in order to derive its Taylor series expansion in the first place, so I'm not sure whether you are satisfied with this answer.
  4. Nov 23, 2011 #3
    I think this is not possible. Maybe your friend considers


    to be definitions of sine and cosine? But sine and cosine are defined to y/r and x/r respectively.
  5. Nov 23, 2011 #4


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    Assuming we're allowed that definition of cosine, then that also works.

    I'm not sure I understand your statement here. Can you clarify?
  6. Nov 23, 2011 #5


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    Non-sense- because you haven't said what x, y, and r are. There are, in fact, many different ways to define sine and cosine. The formulas given above are perfectly good as definitions.

    You could also define cos(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n)!]x^{2n}[/itex]
    and sine(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n+1)!]x^{2n+1}[/itex]

    Or you can define cos(x) as the function satifying y''= -y with initiial conditions y(0)= 1, y'(0)= 0 and define sin(x) as the function satisfying y''= y with initial conditions y(0)= 0, y'(0)= 1.
  7. Nov 23, 2011 #6
    If you draw a right-angled triangle, then x is the adjacent side, y is the opposite side and r is the hypotenuse.
  8. Nov 23, 2011 #7
    Yes but I am considering the geometric definition.

    Isn't the second derivative of sin(x) is -sin(x) ?
  9. Nov 23, 2011 #8


    Staff: Mentor

    Yes. I'm pretty sure HallsOfIvy meant y'' + y = 0 as the differential equation.
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