# What is an antiderivative for $\frac{1}{1+3\cos^{2}x}$?

Define $$f:\mathbb R \rightarrow \mathbb R$$ by $$f(x)=\frac{1}{1+3\cos^{2}x}$$.

Since $$f$$ is continuous, $$f$$ has an antiderivative. That is, there exists some function $$F:\mathbb R \rightarrow \mathbb R$$ such that $$F^{\prime}=f$$.

Using the substitution $$u=\tan x$$ and doing some work, we find that a possible antiderivative of $$f$$ is the function $$g$$ defined by $$g(x)=\frac{1}{2}\tan^{-1}\frac{\tan x}{2}$$.

However, the problem with $$g$$ is that it's not defined at odd multiples of $$\pi/2$$, while the $$F$$ we're seeking is defined on $$\mathbb R$$.

So, what is $$F$$ and how do we find it?

(Or is there a mistake somewhere in my argument above?)

• I won't mark this one as a duplicate since the question statements are rather different, but the issue is exactly the same as the one in this question, which has good* answers taking various approaches, involving a similar integral: math.stackexchange.com/questions/3275770/… (*One of the answers is mine, I make no claims about its goodness.) – Travis Willse Feb 23 at 4:21

For each odd integer $$k$$, define $$f_k:((k-2)\pi/2,k\pi/2) \rightarrow \mathbb R$$ by $$f_k(x)=\frac{1}{1+3\cos^{2}x}$$.

As you've found, an antiderivative of $$f_k$$ is $$g_k:((k-2)\pi/2,k\pi/2) \rightarrow \mathbb R$$ defined by $$g_k(x)=\frac{1}{2}\tan^{-1}\frac{\tan x}{2}$$.

And as you've also correctly stated, we know by the FTC that $$f$$ must have an antiderivative $$F$$.

So, we guess that such a function $$F$$ can be constructed by "moving up or down" the functions $$g_k$$ and "stitching" them together, and also "filling in the holes" at each odd multiple of $$\pi/2$$.

By examining the graphs of each $$g_k$$, we come up with this guess:

For each odd integer $$k$$, define $$F$$ by $$F(x)=\begin{cases} \left(k-1\right)\frac{\pi}{2}, & \text{ for }x=\left(k-2\right)\frac{\pi}{2},\\ \frac{1}{2}\tan^{-1}\frac{\tan x}{2}+k\frac{\pi}{4}, & \text{ for }x\in\left(\left(k-2\right)\frac{\pi}{2},k\frac{\pi}{2}\right)\\ k\frac{\pi}{2}, & \text{ for }x=k\frac{\pi}{2}. \end{cases},$$

To show that the above guess is correct, we can verify that $$F^{\prime}=f$$ (this will involve more work).

Similar questions: 1, 2, 3. See also Jeffrey and Rich (1994, "The Evaluation of Trigonometric Integrals Avoiding Spurious Discontinuities") and Jeffrey (1994, "The Importance of Being Continuous").