Question

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

• A. $0$
• B. $\pi -\frac{{\pi }^2}{3}$
• C. $2\pi -{\pi }^3$
• D. $\frac{7}{2}-2{\pi }^3$

Answer 1

The correct option is A $0$

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