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It is quite easy to prove, that if $f(x)$ is integrable on $[a, b]$ then $F(x)=\int_{a}^{x} f(t)\,dt$ is continuous on $[a, b]$. Also if $f(x)$ is continuous on $[a, b]$ then $F(x)$ is differentiable and $F(x)$ is an antiderivative of $f(x)$. And the antiderivative of the function is continuous by definition.

Now let's consider function:

$$ f(x)=\frac{1}{2\cos^2(x)+\sin^2(x)}$$

It is continuous on $[0, 2\pi]$ but its antiderivative is not:

$$ F(x)=\frac{1}{\sqrt{2}}\,\arctan\left(\frac{\tan(x)}{\sqrt{2}}\right) $$.

It is not continuous at $\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$. So what am I missing? Thanks.

Adrian Keister
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Andrew Popow
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    The point is that a continuous antiderivative exists, but it is a piecewise function if written in the form you wrote, with the constant of integration taking a different value on each interval between odd multiples of $\pi/2$. You can actually construct such an antiderivative by integrating on a definite integral with a variable endpoint, in which case you will see that you have to split the domain of integration up in order to consistently use the substitution you used. – Ian Sep 18 '19 at 18:46
  • Cf. this question and the linked questions there: https://math.stackexchange.com/questions/1356523/what-are-the-restrictions-on-using-substitution-in-integration – Hans Lundmark Sep 18 '19 at 19:29

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