Answer
$$18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + \frac{1}{2}\sqrt {36 - {t^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {36 - {t^2}} dt} \cr
& {\text{substitute }}t = 6\sin \theta ,{\text{ }}dt = 6\cos \theta d\theta \cr
& = \int {\sqrt {36 - 36{{\sin }^2}\theta } \left( {6\cos \theta } \right)d\theta } \cr
& = \int {6\sqrt {1 - {{\sin }^2}\theta } \left( {6\cos \theta } \right)d\theta } \cr
& {\text{pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = 36\int {\sqrt {{{\cos }^2}\theta } \cos \theta d\theta } \cr
& = 36\int {{{\cos }^2}\theta d\theta } \cr
& = 36\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& {\text{integrating}} \cr
& = 36\left( {\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right) + C \cr
& = 18\theta + 9\sin 2\theta \cr
& = 18\theta + 18\sin \theta \cos \theta \cr
& = 18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + 18\left( {\frac{t}{6}} \right)\left( {\frac{{\sqrt {36 - {t^2}} }}{6}} \right) + C \cr
& {\text{simplify}} \cr
& = 18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + \frac{1}{2}\sqrt {36 - {t^2}} + C \cr} $$
