## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{1}{4}x^{2}-\frac{1}{4}\sin(x^{2})\cos(x^{2})+C$ The integrand, $f(x)$ (red graph) behaves as $F'(x)$ so the answer is reasonable. 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.