Answer
$$\int\sqrt x\sin^2(x^{3/2}-1)dx=\frac{x^{3/2}-1}{3}-\frac{\sin(2x^{3/2}-2)}{6}+C$$
Work Step by Step
$$A=\int\sqrt x\sin^2(x^{3/2}-1)dx$$
We set $u=x^{3/2}-1$.
Then $$du=\frac{3}{2}x^{1/2}dx=\frac{3}{2}\sqrt xdx$$
That means $$\sqrt xdx=\frac{2}{3}du$$
Therefore, $$A=\int\sin^2u\frac{2}{3}du=\frac{2}{3}\int\sin^2udu$$
Recall the identity $\sin^2u=\frac{1-\cos2u}{2}$
$$A=\frac{2}{3}\int\frac{1-\cos2u}{2}du=\frac{1}{3}\int(1-\cos2u)du$$ $$A=\frac{1}{3}\Big(u-\frac{1}{2}\sin2u\Big)+C$$ $$A=\frac{u}{3}-\frac{\sin2u}{6}+C$$ $$A=\frac{x^{3/2}-1}{3}-\frac{\sin(2x^{3/2}-2)}{6}+C$$
