Find the value of $\int_{-\infty}^{\infty} \cos^{-1}\left(\cos\left(\frac{24+4 x^2}{4+x^2}\right)\right)dx$.

I tried separating the limits or converting it to a different trigonometric function. Neither of them worked. Please suggest a method to remove the inverse function

  • $\begingroup$ What have you tried? Just posting a homework question won't be that well received on this site I might consider integral-calculator.com for this one. $\endgroup$
    – Nic
    Commented May 16 at 4:38
  • $\begingroup$ I tried removing the inverse function in each separate limit but that was a very long procedure.... i just wanted to know a method to start it. Because since its an inverse it will be different in each part of the limit $\endgroup$ Commented May 16 at 4:40
  • $\begingroup$ I would suggest editing the question and adding a bit more of your attempt @bhargavinarayanan $\endgroup$
    – Sam
    Commented May 16 at 4:48
  • $\begingroup$ $\cos^{-1}(cos(x)) \ne x$ for all $x \in \mathbb{R}$. See my hint below @Hussain-Alqatari $\endgroup$
    – Sam
    Commented May 16 at 4:56
  • $\begingroup$ @Sam, You are right. (Comment deleted). $\endgroup$ Commented May 16 at 4:59

3 Answers 3


Note that $\frac{24+4x^{2}}{4+x^{2}} = \frac{8}{4+x^{2}}+4$ Now, when $x$ goes to infinity or negative of infinity, the expression tends to $4$, and when $x$ goes to $0$, expreession tends to 6.

Thus,the expression lies between $[4,6]$.

$\cos^{-1}\cos(y) = -y+2\pi$ when $y \in [\pi,2\pi]$. Now take $y$ to be the above expression. It's an easy integral to compute now.

The integral reduces to :

$\int_{-\infty}^{\infty} 2\pi -4 - \frac{8}{4+x^{2}} = (2\pi-4)x|_{-\infty}^{\infty} - \frac{1}{2}\tan^{-1}\frac{x}{2}|_{-\infty}^{\infty} $. The last expression is $\frac{-1}{2}(\frac{\pi}{2}-\frac{-\pi}{2}) = -\frac{\pi}{2}$ and the first expression is $\infty$

My computation gives me that the integration is $ \infty$


Adding on to the hint of @Robert, if you are wondering how $\cos ^{-1}(\cos x)$ is evaluated for $x \in \mathbb{R}$:

$\cos ^{-1}(\cos x)=x$ if $0 \leq x \leq \pi$. If $x \in[n \pi,(n+1) \pi]$, then $x-n \pi \in[0, \pi]$. We get, $$ x-n \pi=\cos ^{-1}(\cos (x-n \pi))=\cos ^{-1}\left((-1)^n \cos x\right)= \begin{cases}\cos ^{-1}(\cos x), & n \text { even } \\ \pi-\cos ^{-1}(\cos x), & n \text { odd }\end{cases} $$

Which gives us the result $\cos ^{-1}(\cos x)=\left\{\begin{array}{ll}x-n \pi, & n \text { even } \\ -x+(n+1) \pi, & n \text { odd }\end{array}\right.$.

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Hint: $\cos^{-1}(\cos(t))$ is a piecewise linear function that is $0$ at even multiples of $\pi$ and $\pi$ at odd multiples of $\pi$. $(24+4x^2)/(4+x^2)$ is always between $\pi$ and $2\pi$.

But too bad you didn't have an integrand whose limit as $x \to \pm \infty$ was $0$, giving your improper integral a chance to converge.


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