Hi , after a substitution with x=3 cos u the answer in the text gives the following...see image
I can't see how it tidies up to u cos u
thank you
Let $\displaystyle \displaystyle \begin{align*} x = 3\cos{u} \implies dx = -3\sin{u}\,du \end{align*}$, and noting that when $\displaystyle \displaystyle \begin{align*} x = 0, u = \frac{\pi}{2} \end{align*}$ and when $\displaystyle \displaystyle \begin{align*} x = \frac{3}{2}, u = \frac{\pi}{3} \end{align*}$, the integral becomes
$\displaystyle \displaystyle \begin{align*} \int_0^{\frac{3}{2}}{\frac{x\arccos{\left(\frac{x} {3}\right)}}{\sqrt{9 - x^2}}\,dx} &= \int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{3\cos{u }\arccos{\left(\frac{3\cos{u}}{3}\right)}}{\sqrt{9 - \left(3\cos{u}\right)^2}}\cdot -3\sin{u}\,du} \\ &= -9\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{u\cos{ u}\sin{u}}{\sqrt{9 - 9\cos^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{ u}\sin{u}}{\sqrt{9\sin^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{ u}\sin{u}}{3\sin{u}}\,du} \\ &= 3\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{u\cos{u}\,du } \end{align*}$
which you can now solve using Integration by Parts