TITLE: Using an integration technique:$ \int^b_a \left[f(x)+f^{-1}(x)\right]dx=b^2-a^2$
QUESTION [6 upvotes]: I read about this integration technique on quora:


If $a,b$ are fixed points of $f$, then $$ \int^b_a \left[f(x)+f^{-1}(x)\right]dx=b^2-a^2$$


Apparently it was used in the final of the 2013 MIT Integration bee but I can't find that question anywhere..
REPHRASING QUESTION: Could someone show me some integrals that can be cleverly solved with this technique?

REPLY [2 votes]: Problem: Evaluate
$$\int\limits_{0}^{1}{\arcsin x\,dx}.$$
Solution: Let $f(x) = \sin\left(\frac{\pi}{2}x\right)$. Then $0$ and $1$ are fixed points of $f$, and $f$ is increasing (and hence invertible) on $(0,1)$. Since $f^{-1}(x) = \frac{2}{\pi}\arcsin x$, we have
$$ \int\limits_{0}^{1}{\sin\left(\frac{\pi}{2}x\right) + \frac{2}{\pi}\arcsin x\,dx} = 1.$$
Since
$$ \int\limits_{0}^{1}{\sin\left(\frac{\pi}{2}x\right)\,dx} = \frac{2}{\pi}\int\limits_{0}^{\pi/2}{\sin u\,du} = \frac{2}{\pi} $$
it follows that
$$ \frac{2}{\pi}\int\limits_{0}^{1}{\arcsin x\,dx} = 1-\frac{2}{\pi} \implies \int\limits_{0}^{1}{\arcsin x\,dx} = \boxed{\frac{\pi}{2}-1}. $$
Of course, $\arcsin x$ does have an elementary antiderivative which can be found using integration by parts, but this method seems faster to me.