Answer
$\displaystyle \frac{1}{4}x^{2}-\frac{1}{4}\sin(x^{2})\cos(x^{2})+C$
The integrand, $f(x) $
(red graph) behaves as $F'(x)$ so the answer is reasonable.
Work Step by Step
Substitute $\left[\begin{array}{ll}
u=x^{2} & \\
du=2xdx &
\end{array}\right]$ .
$I=\displaystyle \int\sin^{2}(x^{2})\cdot(xdx)=\int\sin^{2}u(\frac{1}{2}du)$
Use the double angle identity $\cos 2u=1-2\sin^{2}u$
$=\displaystyle \frac{1}{2}\int\frac{1}{2}(1-\cos 2u)du$
$=\displaystyle \frac{1}{4}(u-\frac{1}{2}\sin 2u)+C$
$=\displaystyle \frac{1}{4}u-\frac{1}{4}(\frac{1}{2}\cdot 2\sin u\cos u)+C$
... bring back x,
$=\displaystyle \frac{1}{4}x^{2}-\frac{1}{4}\sin(x^{2})\cos(x^{2})+C$
Graphing, the integrand $f(x)=x\sin^{2}(x^{2})$
should behave as $F'(x)$, where $F(x)=\displaystyle \frac{1}{4}x^{2}-\frac{1}{4}\sin(x^{2})\cos(x^{2})$
When $ F$ decreases (from the left to the origin), $f$ is negative.
When $ F$ increases (from the origin onwards), $f$ is positive.
At the minimum point of F (the origin), f is zero.
At inflection points of F, f is zero.
$f(x) $(red graph) behaves as $F'(x)$ so the answer is reasonable.
