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I'm interested in why $$\int_0^{\pi/2} \sin x\,dx = 1.$$ I know how to do the integral the conventional way but am more interested in what makes radians special for this problem. If we instead compute $$\int_{0}^{90} \sin x^\circ\,dx,$$ we won't get $1$ as the answer.

What about the definition of radians makes this integral evaluate to $1$? I'm looking for an intuitive (presumably geometric) explanation.

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    If you measure in degrees instead of radians, the derivative of $\cos(x)$ is not $-\sin(x)$; you're off by a constant (think the graph: instead of having a period of $2\pi$, it has a period of $360$, almost 60 times longer). So $\cos(x)$ is no longer an antiderivative of $\sin(x)$. If you do the integral with the correct antiderivative, you *wll* get $1$.2011-02-13
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    Mathematica says (and I'm inclined to agree) that $\int_0^{\pi/2} \sin x\,dx =\int_{0^\circ}^{90^\circ} \sin x\,dx= 1$, whereas $\int_{0}^{90} \sin x^\circ\,dx=\frac{1}{1^\circ}=\frac{180}{\pi}$ (though this is more of an issue of notation than actual substance).2011-02-13
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    @Isaac, you're right that that does make more sense. I'll change my question (though I'm not sure it was particularly unclear originally).2011-02-13
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    @Arturo: Yes, but _why_ is the derivative of $\cos(x)$ equal to $-\sin(x)$ when $x$ is measured in radians?2011-02-13
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    @Ben: Because $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$when $x$ is measured in radians. Basically, because radians give the arclength parametrization of the unit circle (one radian yields one unit of arc length), and degrees do not. Radians are the "normalized" way of measuring angles, so that pesky proportionality constants all become $1$ when you use radians.2011-02-13
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    @Ben: The question, "Why is the derivative of $\cos x$ equal to $- \sin x$ when $x$ is measured in radians?" was answered at this math.SE question: http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx.2011-02-13

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