Question

$\text{Evaluate}:{\int }_{-\pi }^{\pi }(1-x^2)\sin x{\cos }^{2}xdx$

Answer 1

We know that
${\int }_{-a}^{a}f\left(x\right)dx=\{\begin{array}{c}\begin{array}{ccc}& 2{\int }_0^{a}f\left(x\right)dx,& \text{if f(x) is even}\\ & 0,& \text{if f(x) is odd}\end{array}\end{array}$

Here, $f\left(x\right)=(1-x^2)\sin x{\cos }^{2}x$
$f(-x)=(1-(-x{)}^2\left)\sin \right(-x\left){\cos }^2\right(-x)$
$=-(1-x^2)\sin x{\cos }^{2}x=-f\left(x\right)$
$\Rightarrow f\left(x\right)$ is an odd function.
$\therefore {\int }_{-\pi }^{\pi }(1-x^2)\sin x{\cos }^{2}xdx=0$