# ostensibly discontinuous antiderivative

 $\displaystyle x\;\mapsto\;\frac{1}{5-3\cos{x}}$ (1)

is continuous   for any $x$ (the denominator is always positive) and therefore it has an antiderivative, defined for all $x$.  Using the universal trigonometric substitution  $\cos{x}\;:=\;\frac{1\!-\!t^{2}}{1\!+\!t^{2}},\qquad dx\;=\;\frac{2dt}{1\!+\!t^% {2}},\qquad t\,=\,\tan\frac{x}{2},$

we obtain

 $5-3\cos{x}\;=\;\frac{5(1\!+\!t^{2})-3(1\!-\!t^{2})}{1\!+\!t^{2}}\;=\;\frac{2(1% \!+\!4t^{2})}{1\!+\!t^{2}},$

whence

 $\int\!\frac{dx}{5-3\cos{x}}\;=\;\int\!\frac{dt}{1\!+\!4t^{2}}\,=\,\frac{1}{2}% \arctan 2t+C\,=\,\frac{1}{2}\arctan\!\left(2\tan\frac{x}{2}\right)+C.$

This result is not defined in the odd multiples of $\pi$, and it seems that the function  (1) does not have a continuous antiderivative.

However, one can check that the function

 $\displaystyle x\;\mapsto\;\frac{x}{4}+\frac{1}{2}\arctan\frac{\sin{x}}{3-\cos{% x}}+C$ (2)

is everywhere continuous and has as its derivative  the function (1); one has

 $\left|\frac{\sin{x}}{3-\cos{x}}\right|\;\leqq\;\frac{1}{3\!-\!1}\;=\;\frac{1}{% 2}\;<\;\frac{\pi}{2}.$

## References

• 1 Ernst Lindelöf: Johdatus korkeampaan analyysiin. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
Title ostensibly discontinuous antiderivative OstensiblyDiscontinuousAntiderivative 2013-03-22 18:37:08 2013-03-22 18:37:08 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 26A36 CyclometricFunctions