The value of ${\int }_{-\pi }^{\pi } \left(1-x^2\right)\sin x{\cos }^{2}xdx$ is

Let $I={\int }_{-\pi }^{\pi } \left(1-x^2\right)\sin x{\cos }^{2}xdx$

As we know

$\begin{array}{l}{\int }_{-a}^a f\left(x\right)dx=2{\int }_0^a f\left(x\right)dx ;\text{ if }f(-x)=f\left(x\right)\\ =0 ; \text{ if }f(-x)=-f\left(x\right)\end{array}$

Consider $f\left(x\right)=\left(1-x^2\right)\sin x{\cos }^{2}x$

Put $x=-x$, we get

$f(-x)=\left[1-(-x{)}^2\right]\sin (-x){\cos }^2(-x)$

$\begin{array}{l}=\left(1-x^2\right)(-\sin x){\cos }^{2}x\\ =-\sin x{\cos }^{2}x\left(1-x^2\right)\\ =-f\left(x\right)\end{array}$

Thus f(x) is an odd function

$\therefore {\int }_{-\pi }^{\pi } \left(1-x^2\right)\sin x{\cos }^{2}xdx=0$