Q. The value of $\int_{- \pi}^{\pi}{\left( 1 - x^{2} \right)\sin{x\cos^{2}{x\ dx\ }}}$is
A) 0
B) $\pi - \frac{\pi^{3}}{3}$
C) $2\pi - \pi^{3}$
D) $\frac{7}{2} - 2\pi^{3}$
Explanation
(a)
Let $f(x) = \left( 1 - x^{2} \right)\sin{x\cos^{2}x}$
$f( - x) = \left\lbrack 1 - ( - x)^{2} \right\rbrack\left\lbrack \sin( - x) \right\rbrack\cos^{2}{( - x)}$
$= - \left( 1 - x^{2} \right)\sin{x\cos^{2}{x = - f(x)}}$
$\Rightarrow f(x)$ is an odd function.
$\therefore\int_{- \pi}^{\pi}{\left( 1 - x^{2} \right)\sin{x\cos^{2}{x\ dx = 0}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.