By letting $u = 2x$ and $t = \tan \frac{u}{2}$, I found the continuous antiderivative of the function to be:

$$\int \frac{1}{1+\cos^2 x}dx\\= \int \frac{2}{3+\cos2x} dx\\ = \int \frac{1}{3+\cos u}du \\=\int \frac{\frac{2}{1+t^2}}{3+\frac{1-t^2}{1+t^2}}dt\\= \int\frac{1}{2+t^2}dt \\= \frac{1}{\sqrt2}\arctan\left(\frac{\tan x}{\sqrt2}\right) + \frac{\pi}{\sqrt2} \left\lfloor \frac{x + \frac{\pi}{2} }{\pi} \right\rfloor + C $$

(I graphically deduced the floor function bit as I am not familiar with its algebra.)

However, GeoGebra (notably not wolfram) does it better. It states, without the floor function, that the continuous antiderivative is also:

$$ \frac{x}{\sqrt2} + \frac{1}{\sqrt2} \arctan\left( \frac{(1-\sqrt2)\sin 2x}{(\sqrt2 -1)\cos2x +\sqrt2 + 1}\right) + C$$

enter image description here

How did GeoGebra accomplish such a feat? And how can I prove and apply such ingenuity?

  • $\begingroup$ Can you please show your steps? $\endgroup$ – Gibbs Jun 3 '18 at 18:51
  • $\begingroup$ Should I add it to the question or post in here? But.. why did you remove the parentheses of my trig functions :( I liked those, but thanks for making floor function big. $\endgroup$ – Mint Jun 3 '18 at 18:57
  • $\begingroup$ Adding the steps in the question is fine. Some of the parentheses were not necessary. The expressions $\cos^2 (x), \tan (x)$ are the same as $\cos^2 x, \tan x$, etc. $\endgroup$ – Gibbs Jun 3 '18 at 19:00
  • 2
    $\begingroup$ Related articles: jstor.org/stable/2690852 doi.org/10.1145/174603.174409 $\endgroup$ – StayHomeSaveLives Jun 4 '18 at 14:59

One important thing is while doing u-substitution, the substitution has to be injective. When we substitute $u=\tan x$ in samjoe’s answer, $\tan x$ is not injective on the whole real line, but is injective in intervals of length $\pi$. That’s why you get the right ‘behavior’ only within intervals but not between them.

I believe that the Geogebra’s answer can be derived by noting that $$\frac1{\pi}(arctan(\cot(\pi x))+\pi x-\pi/2)$$ behaves exactly the same as a floor function.

Use also the summation formula for arctan: $$arctan (u)+arctan (v)=arctan(\frac{u+v}{1-uv})$$


To demonstrate the importance of injectivity of substitutions, consider the integral $$\int^1_{0}xdx$$ which equals $\frac12$.

If we substitute in $u=x^2-x$, we obtain something like $$\int^0_0 \cdots du=0$$

What caused the paradox is $x^2-x$ is not injective in the interval $[0,1]$.

Similarly, there is nothing wrong for $$\int^x_k \frac1{1+\cos ^2x}dx=\int^x_k\frac{\sec^2x}{2+\tan^2x}dx=^{u=\tan x}\int^{arctan(x)}_{arctan(k)}\frac{du}{2+u^2}=\frac1{\sqrt2}arctan(\frac{\tan x}{\sqrt2})+C$$ as long as $\tan x$ is injective in the interval $[k,x]$. If the injectivity is not achieved in the interval, $C$ would change when $x$ goes from an injective interval of $\tan x$ to another.

This agrees with what the OP observed: the floor function thing is a constant in each injective interval of $\tan x$, and changes when going across the intervals. You may consider, the floor function thing is part of $C$.

The choice of $k$ is arbitrary. But when we try to find an antiderivative for all $x$ while $k$ remains fixed, it is impossible to always achieve the injectivity in $[k,x]$. As a trade off, we need to add a floor function to compensate for the silent change of $C$.


@samjoe derived the antiderivative $$\frac{\pi}{\sqrt2}\left \lfloor\frac{x+\pi/2}{\pi}\right\rfloor + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) $$

By noting $$\lfloor x\rfloor=\frac1{\pi}(arctan(\cot(\pi x))+\pi x-\pi/2)$$, the above expression can be rewritten to $$\frac{x}{\sqrt2}+\frac{arctan(-\tan x)}{\sqrt2}+\frac{arctan(\frac{\tan x}{\sqrt2})}{\sqrt2}$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}({arctan(-\tan x)}+arctan(\frac{\tan x}{\sqrt2}))$$ By the summation formula stated above $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{-\tan x+\frac{\tan x}{\sqrt2}}{1+\frac{\tan^2x}{\sqrt2}})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{\tan x(1-\sqrt2)}{\sqrt2+\tan^2x+1-1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{\tan x(1-\sqrt2)}{\sqrt2+\sec^2x-1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin x\cos x}{(\sqrt2-1)\cos^2 x+1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)2\sin x\cos x}{(\sqrt2-1)(2\cos^2 x)+2})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin 2x}{(\sqrt2-1)(\cos 2x +1)+2})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin 2x}{(\sqrt2-1)\cos 2x+\sqrt2+1})$$ which is exactly what we want.

  • 1
    $\begingroup$ Nice manipulation +1 :), especially that floor equivalent. But I am wondering how can we find this without resorting to that floor. $\endgroup$ – SJ. Jun 4 '18 at 13:13
  • $\begingroup$ @samjoe I wonder too. $\endgroup$ – Szeto Jun 4 '18 at 13:35
  • $\begingroup$ Wow, thank you! Any tips on finding alternate forms of the floor function? $\endgroup$ – Mint Jun 4 '18 at 15:07
  • 1
    $\begingroup$ @jiaminglimjm I discovered this form simply by playing around with functions.:) $\endgroup$ – Szeto Jun 4 '18 at 22:19

With Floor Function

Let $(1+\cos^2x )^{-1} = f(x)$. Now as you found,

$$\int \frac{dx}{1+\cos^2 x} = \int \frac{\sec^2 x }{2+\tan^2 x} dx = \frac{1}{\sqrt{2} } \arctan\left(\frac{\tan x}{\sqrt 2}\right)$$

The issue is that integral of a continuous function should be continuous. The one we found is discontinuous at all odd multiples of $\pi/2$. Lets analyse for $x\in [\tfrac{(2k-1)\pi}{2}, \tfrac{(2k+1 ) \pi}{2}]$. Then

$$\begin{align} \int_{0}^{x} f(t) dt &= \int_{0}^{\pi/2}f(t) dt+\int_{\pi/2}^{3\pi/2}f(t) dt ... \int_{(2k-1)\pi/2}^{x}f(t) dt \\ &= \frac{\pi k}{\sqrt2} + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) \\ \end{align}$$

Now since $x\in [\tfrac{(2k-1)\pi}{2}, \tfrac{(2k+1 ) \pi}{2}], $ then $x+\pi/2 \in [k\pi, (k+1)\pi]$ and so $\frac{x+\pi/2}{\pi} \in [k, k+1]$ so that $\lfloor\frac{x+\pi/2}{\pi}\rfloor = k$. Substituting in above equation gives:

$$\int_{0}^{x} f(t) dt =\frac{\pi}{\sqrt2}\left \lfloor\frac{x+\pi/2}{\pi}\right\rfloor + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) $$

Without Floor Function

Couldn't do this one, will add if I find one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.