# Finding the antiderivatives of a trigonometric function in two different intervals

I wanted to find the antiderivatives of the function $$f(x)=\frac{1}{sin(x)+cos(x)+2}$$ in $$[0,\pi[$$ first and then in $$[0,2\pi]$$. Now, as long as the first set is concerned, I used Weierstrass's substitution to find that $$\int f(x)dx=\sqrt2 \cdot arctan \left(\frac{tan(x/2)+1}{\sqrt2}\right)+c$$ This result seems fine to me since the antiderivatives of $$f$$ are well defined in $$[0,\pi[$$. But when we consider $$[0,2\pi]$$, the antiderivatives are not well defined since $$F(x)=\sqrt2 \cdot arctan \left(\frac{tan(x/2)+1}{\sqrt2}\right)$$ is not continuous at $$x=\pi$$, so I ended up saying that in this case $$\int f(x)dx=\emptyset$$ I think the reason why this is true (please correct me if I'm wrong) is that, since our function is defined on an interval ($$[0,2\pi]$$ in this case), its antiderivatives must differ by a constant everywhere in $$[0,2\pi]$$, but this obviously doesn't happen when $$x=\pi$$, i.e taken two antiderivatives, $$F$$ and $$G$$, we couldn't write $$G(x)=F(x)+c \ \forall x\in[0,2\pi]$$, because it wouldn't make any sense when $$x=\pi$$

the antiderivative you obtain is reliable only in the interval $$-\pi < x < \pi$$ since the Weierstrass substitution is defined only for this particular interval. Indeed the function $$f(x) = \frac{1}{\sin(x)+\cos(x)+2}$$ is continuous and so it has a continuous antiderivative on all $$\mathbb{R}$$.

That being said having the right antiderivative in $$(-\pi,\pi)$$ is enough to create the antiderivative on all $$\mathbb{R}$$ because the function has period $$2\pi$$.

We will use $$f(x-\pi)$$ so that the formulation of $$\int_0^{x} f(x)dx$$ is easier. The indefinite integral is

$$\int f(x-\pi)dx=-\sqrt2 \arctan \left(\frac{1-3\tan(\frac{x}{2})}{\sqrt2}\right), -\pi < x < \pi$$

and so in general

\begin{align} \newcommand{\floor}[1]{\lfloor #1 \rfloor}\int_0^{x} f(x)dx &= \floor{\frac{x}{2\pi}}\int_{-\pi}^{\pi} f(x-\pi)dx+ \int_{-\pi}^{x-\pi-2\pi\floor{\frac{x}{2\pi}}}f(x-\pi)dx \\ & = \floor{\frac{x}{2\pi}}\sqrt{2}\pi+ \int_{-\pi}^{x-\pi-2\pi\floor{\frac{x}{2\pi}}}f(x-\pi)dx \\ & = \floor{\frac{x}{2\pi}}\sqrt{2}\pi - \sqrt2 \arctan \left(\frac{1-3\tan(\frac{x-\pi-2\pi\floor{\frac{x}{2\pi}}}{2})}{\sqrt2}\right)+\frac{\pi}{\sqrt{2}} \end{align}

As you can see on Wolfram this is a continuous function as expected.

There is one final concern: because of the Weirstrass substitution is defined on $$(-\pi,\pi)$$ you will find some undefined points inside the function above, specifically $$x=2n\pi$$ for $$n \in \mathbb{N}$$, but because the antiderivative is continuous you can define these points as the limit of the function which gives you

$$F(x) = \begin{cases} \sqrt{2} n \pi & \mbox{if } x = 2n\pi \\ \floor{\frac{x}{2\pi}}\sqrt{2}\pi - \sqrt2 \arctan \left(\frac{1-3\tan(\frac{x-\pi-2\pi\floor{\frac{x}{2\pi}}}{2})}{\sqrt2}\right)+\frac{\pi}{\sqrt{2}} & \mbox{otherwise } \end{cases}$$

• Is the floor function differentiable everywhere in $[0,2\pi]$? Dec 30, 2021 at 16:50
• That's a kind way to say that now I understand the problem even less than before Dec 30, 2021 at 16:58
• Moreover, I'm not interested in finding the antiderivatives in $\mathbb{R}$, but just in $[0,2\pi]$ (if there's any) Dec 30, 2021 at 17:11
• the antiderivative is well defined on all $\mathbb{R}$ (that is because $f$ is bounded and continuous so it is sure that the antiderivative exists) and so also on $[0,2\pi]$, you need to be careful though because there are points where the antiderivative I wrote is not well defined because of the Weierstrass substitution but because you know that the antiderivative is a continuous function you can substitute the limit to that specific value e.g. take for example $x=0$ in the last expression, it is undefined but you have the easy fix that instead you can consider $lim_{x \to 0}$ of it Dec 30, 2021 at 17:18
• I found your question interesting and so I addressed it in a general way, also look at here for the existance of the antiderivative...say to me if something is not clear :-), [ Dec 30, 2021 at 17:25

Thanks to @Tortar's answer I think I found out what was missing in my reasoning and now I'm going to explain it. To begin with, we want to find the antiderivatives of $$f$$ in $$[0,\pi[$$, and we determined that in this case $$\int f(x)dx=\sqrt2 \cdot arctan\left(\frac{tan(x/2)+1}{\sqrt2}\right)+c$$ So far, so good. Now, in $$[0,2\pi]$$ our $$f$$ is continuous and so we expect to find antiderivatives due to Torricelli's theorem. Let $$F(x)=\sqrt2 \cdot arctan\left(\frac{tan(x/2)+1}{\sqrt2}\right)+c$$ and $$G$$ be a function such that $$G'(x)=f(x) \ \forall x \in[0,2\pi]$$. From what I said until now, it follows that $$G$$ is an antiderivative of $$f$$ in $$[0,\pi[$$, so it must differ to $$F$$ by a constant (the same argument is valid in $$]\pi,2\pi])$$. So this should give us an hint about the fact that the antiderivatives in $$[0,2\pi]$$ are "similar" to $$F(x)$$, which has to be made continuous at $$x=\pi$$. So let us redefine $$F$$ in the following way $$F(x)=\begin{cases}\sqrt2 \cdot arctan\left(\frac{tan(x/2)+1}{\sqrt2}\right)+c, \ \ \ x\in[0,\pi[ \\ \frac{\pi}{\sqrt2}+c, \ \ \ x=\pi \\\sqrt2 \cdot arctan\left(\frac{tan(x/2)+1}{\sqrt2}\right)+\sqrt2\pi+c, \ \ \ x\in]\pi,2\pi]\end{cases}$$ Which is indeed (or at least, it should be) an antiderivative of $$f$$ in $$[0,2\pi]$$

• you nailed it, that is the antiderivative for $[0,2\pi]$! Dec 31, 2021 at 12:49
• Thank you so much! I upvoted your answer as well Dec 31, 2021 at 13:11