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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

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In summary, the conversation discusses the possibility of finding the integral of cosine without using the fact that the derivative of sine is cosine. Suggestions include using the Taylor series expansion of cosine and defining sine and cosine in different ways, such as using complex numbers or geometric definitions. However, there is some disagreement on what constitutes a valid definition of sine and cosine, and it is unclear if it is possible to find the integral without using this derivative relationship.

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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

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- #2

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ahaanomegas said:

Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

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.

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ahaanomegas said:

Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

I think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine? But sine and cosine are defined to y/r and x/r respectively.

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dimension10 said:I think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine?

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

dimension10 said:But sine and cosine are defined to y/r and x/r respectively.

I'm not sure I understand your statement here. Can you clarify?

- #5

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Non-sense- because you haven't said what x, y, and rdimension10 said:I think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine? But sine and cosine are defined to y/r and x/r respectively.

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.

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cepheid said:I'm not sure I understand your statement here. Can you clarify?

HallsofIvy said:You haven't said what x, y, and rare.

If you draw a right-angled triangle, then x is the adjacent side, y is the opposite side and r is the hypotenuse.

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HallsofIvy said:Non-sense- because you haven't said what x, y, and rare. There are, in fact, many different ways todefinesine 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]

Yes but I am considering the geometric definition.

HallsofIvy said:define sin(x) as the function satisfying y''= y with initial conditions y(0)= 0, y'(0)= 1.

Isn't the second derivative of sin(x) is -sin(x) ?

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Yes. I'm pretty sure HallsOfIvy meant y'' + y = 0 as the differential equation.dimension10 said:Isn't the second derivative of sin(x) is -sin(x) ?

A basic integral without basic concept refers to the process of calculating the area under a curve without using the fundamental concept of integration. This method relies on basic geometric shapes, such as rectangles, to approximate the area instead of using the integration formula.

Using a basic integral without basic concept can be useful when the function being integrated is too complex or difficult to integrate using traditional methods. It can also be a quicker and simpler approach for finding an estimate of the area under a curve.

While this method can provide a rough estimate of the area under a curve, it is not as accurate as using the traditional integration formula. It also cannot be used for functions that have discontinuities or curves that are not easily approximated by geometric shapes.

To calculate a basic integral without basic concept, you first divide the x-axis into equal intervals, then use the function to find the height of each interval. Next, you approximate the area under the curve by multiplying the height of each interval by the width of the interval, and summing up all the areas.

No, a basic integral without basic concept can only be used for continuous functions that can be approximated by geometric shapes. It cannot be used for functions with discontinuities or those that cannot be easily approximated by shapes such as circles or ellipses.

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