Answer
$${\text{part a) }} - 2\cos \sqrt x + C{\text{ part b) }}\frac{3}{4}\sqrt {4{x^2} + 5} + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {\frac{1}{{\sqrt x }}\sin \sqrt x } dx \cr
& {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr
& = \int {\frac{1}{{\sqrt x }}\sin \sqrt x } dx = \int {2\sin u} du \cr
& {\text{find antiderivative}} \cr
& = - 2\cos u + C \cr
& {\text{write in terms of }}x \cr
& = - 2\cos \sqrt x + C \cr
& \cr
& {\text{part b}} \cr
& \cr
& \int {\frac{{3xdx}}{{\sqrt {4{x^2} + 5} }}} \cr
& {\text{substitute }}u = 4{x^2} + 5,{\text{ }}du = 8xdx \cr
& = \int {\frac{{3xdx}}{{\sqrt {4{x^2} + 5} }}} = 3\int {\frac{{1/8du}}{{\sqrt u }}} \cr
& = \frac{3}{8}\int {{u^{ - 1/2}}} du \cr
& {\text{find antiderivative}} \cr
& = \frac{3}{8}\frac{{{u^{1/2}}}}{{1/2}} + C \cr
& = \frac{3}{4}{u^{1/2}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{3}{4}\sqrt {4{x^2} + 5} + C \cr} $$
