if $ f(x)=x+\cos x $ then find $ \int_0^\pi (f^{-1}(x))\text{dx} $?

Question

I would be interest to show :

if $ f(x)=x+\cos x $ then find $ \int_0^\pi (f^{-1}(x))\text{dx} $ ?

my second question that's make me a problem is that :

what is :$ f^{-1}(\pi) $ ?

I would be interest for any replies or any comments .

Answer 1

Let $a,b \in \mathbb{R}$ be such that $f(a)=0$ and $f(b)=\pi$. Then $f$ is monotonically increasing and (hence) one-one on the interval $[a,b]$. So to get the integral consider the rectangle formed by $[0,b] \times [0,\pi]$. Then \begin{align*} \int_{0}^{\pi}f^{-1}(x) \, dx & = \pi b- \int_{a}^{b} x + \cos x \, dx\\ & = \pi b- \left(\frac{b^2-a^2}{2}\right) + \left(\sin b -\sin a\right)\\ \end{align*} So all you need is the value of $a$ and $b$ which can be obtained numerically.

Answer 2

There is something questionable in the wording (about the bounds of the integral). So, two interpretations are presented below :

Answer 3

$$\displaystyle \int_{a}^{b}f(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$

$$\displaystyle \int_{a}^{b}(x+\cos x)dx+\int_{0}^{\pi}f^{-1}(x)dx=b\pi$$ $$ f(b)=\pi\Rightarrow b+\cos b=\pi\Rightarrow b\approx 3.88$$ $$f(a)=0\Rightarrow a+\cos a=0\Rightarrow a\approx -0.74$$ $$\displaystyle \int_{0}^{\pi}f^{-1}(x)dx=b\pi-\frac{b^{2}-a^{2}}{2}+(\sin( b)-\sin(a))$$