Answer
$$\int x^{1/2}\sin(x^{3/2}+1)dx=-\frac{2}{3}\cos(x^{3/2}+1)+C$$
Work Step by Step
$$A=\int x^{1/2}\sin(x^{3/2}+1)dx$$
We set $u=x^{3/2}+1$
Then $$du=\frac{3}{2}x^{1/2}dx$$
That means $$x^{1/2}dx=\frac{2}{3}du$$
Therefore, $$A=\frac{2}{3}\int\sin udu=-\frac{2}{3}\cos u+C$$ $$A=-\frac{2}{3}\cos(x^{3/2}+1)+C$$
