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

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

• 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.
– Nic
Commented May 16 at 4:38
• 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 Commented May 16 at 4:40
• I would suggest editing the question and adding a bit more of your attempt @bhargavinarayanan
– Sam
Commented May 16 at 4:48
• $\cos^{-1}(cos(x)) \ne x$ for all $x \in \mathbb{R}$. See my hint below @Hussain-Alqatari
– Sam
Commented May 16 at 4:56
• @Sam, You are right. (Comment deleted). Commented May 16 at 4:59

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

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.