Solution:

$I=\int_{0}^{1} x^{5} \sin ^{-1} x d x$

Put $\sin ^{-1} x=t \quad \therefore x=\sin t d x=\cos t d t$

When $x=0, t=0 \quad$ when $x=1, t=\pi / 2$

$I=\int_{0}^{\pi / 2} \sin ^{5} t . t . \cos t d t=\int_{0}^{\pi / 2} t\left(\sin ^{5} t . \cos t\right) d t$

Integrating by parts,

$I=\left[t \cdot \frac{\sin ^{6} x}{6}\right]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \frac{\sin ^{6} x}{6} .1 . d t$

$I=\left(\frac{\pi}{2} \cdot \frac{1}{6}-0\right)-\frac{1}{6} \cdot \frac{5.3 .1}{6.4 .2} \cdot \frac{\pi}{2}$

$I=\frac{\pi}{12}-\frac{5 \pi}{192}$

$\therefore I=\frac{11 \pi}{192}$