# Arbitrary Ranges

In my last post I discussed the idea that the ranges of the inverse trigonometric functions are chosen somewhat arbitrarily. For good reasons, the ranges always include the first quadrant and the adjoining quadrant (II or IV) where the function is negative. If possible, the range is also chosen to be continuous. Still the choices are arbitrary.

I discussed the range of the inverse tangent function in relation to the value of the improper integral $\int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}$. I noted that if we used some other continuous range for the inverse tangent that the result of this or any other definite integral of this function gives the same value. Thus a range for the inverse tangent of $\left( -\tfrac{\pi }{2},\tfrac{\pi }{2} \right),\left( \tfrac{\pi }{2},\tfrac{3\pi }{2} \right),\left( \tfrac{3\pi }{2},\tfrac{5\pi }{2} \right)$, etc. will give the same result.

For antiderivatives involving the inverse tangent or inverse cotangent this is true, but what about the other inverse trigonometric functions?

When evaluating the difference between two values as one does when evaluating a definite integral any range which results in a graph “parallel” to the graph over the commonly accepted range gives the same value.

However, for an integral requiring the inverse sine, if we use the range $\left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]$, $\displaystyle \int_{0}^{1/2}{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left. {{\sin }^{-1}}\left( x \right) \right|_{0}^{1/2}$ $={{\sin }^{-1}}\left( \tfrac{1}{2} \right)-{{\sin }^{-1}}\left( 0 \right)=\tfrac{5\pi }{6}-\pi =-\tfrac{\pi }{6}$

Indicating that a region above the x-axis has a negative area!

So $\left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]$ is not a good choice. We could use other ranges for the inverse sine function but they would have to be such that they result in inverse sine graphs “parallel” to the usual graph. So we could use $\left( \tfrac{3\pi }{2},\tfrac{5\pi }{2} \right)$ or $\left( \tfrac{7\pi }{2},\tfrac{9\pi }{2} \right)$, but not $\left( \tfrac{5\pi }{2},\tfrac{7\pi }{2} \right)$.

The same problem arises with the inverse cosine, the inverse secant, and the inverse cosecant.

It is best to stick with the commonly accepted ranges. Still, going off on tangents often helps sharpen a student’s understanding.

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