Originally Posted by
Soroban Hello, scorpion007!
We could have tried: .$\displaystyle \int^a_0\cos^{-1}\left(\frac{2x-a}{a}\right)\,dx$
. . but it is far easier to integrate with respect to $\displaystyle y.$
We have: .$\displaystyle y \;= \;\frac{b}{\pi}\cos^{-1}\left(\frac{2x-a}{a}\right)\quad\Rightarrow\quad\frac{\pi}{b}y\;= \;\cos^{-1}\left(\frac{2x - a}{a}\right)$ . $\displaystyle \quad\Rightarrow\quad\frac{2x-a}{a}\;=\;\cos\left(\frac{\pi}{b}y\right) $$\displaystyle
$
. . $\displaystyle 2x - a \;=\;a\cos\left(\frac{\pi}{b}y\right)\quad\Rightar row\quad 2x \;=\;a\cos\left(\frac{\pi}{b}y\right) + a \quad\Rightarrow$ . $\displaystyle x\;=\;\frac{a}{2}\left[\cos\left(\frac{\pi}{b}y\right)+1\right]$
So we have: .$\displaystyle \frac{a}{2}\int^b_0\left[\cos\left(\frac{\pi}{b}y\right) + 1\right]\,dy \;= \;\frac{a}{2}\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}y\right) + y\right]^b_0$
. . $\displaystyle = \;\frac{a}{2}\left\{\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}\cdot b\right) + b\right] - \underbrace{\left[\frac{b}{\pi}\sin(0) + 0\right]}_{0}\right\} $
. . $\displaystyle = \;\frac{a}{2}\left[\frac{b}{\pi}\underbrace{\sin(\pi)}_0 + b\right] \;=\;\frac{ab}{2}$ . . . answer (c)