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Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:
:<math>\begin{align}
\arcsin(x) &{}= \int_0^x \frac{1}{\sqrt{1 \- z^2}} \, dz \; , & |x| &{} \leq 1\\
\arccos(x) &{}= \int_x^1 \frac{1}{\sqrt{1 \- z^2}} \, dz \; , & |x| &{} \leq 1\\
\arctan(x) &{}= \int_0^x \frac{1}{z^2 + 1} \, dz \; ,\\
\arccot(x) &{}= \int_x^\infty \frac{1}{z^2 + 1} \, dz \; ,\\
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