= Solution
For the <indefinite integral>, substitute $u=x^2$, so $x^3dx=u\,du/2$. An <integration by parts> then gives
$$
\int x^3\cos(x^2)dx=\frac12\int u\cos u\,du
=\boxed{\frac12\big[x^2\sin(x^2)+\cos(x^2)\big]+C.}
$$
For the <definite integral>, the original PDF's upper limit is $\sqrt{\pi/2}$, so the corresponding upper value of $u$ is $\pi/2$. Therefore
$$
\int_0^{\sqrt{\pi/2}}x^3\cos(x^2)dx
=\frac12\left[\frac\pi2\sin\frac\pi2+\cos\frac\pi2-1\right]
=\boxed{\frac{\pi-2}{4}.}
$$
Both requested integrals are thus evaluated with the same <antiderivative>.
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