Q99 / 180 · KCET 2024 · Mathematics

$\int_{-\pi}^\pi\left(1-x^2\right) \sin x \cdot \cos ^2 x d x$ is

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

Solution

$\because$ Let $f(x)=\left(1-x^2\right) \sin x \cdot \cos ^2 x$ $f(-x)=\left(1-(-x)^2\right) \sin (-x) \cdot \cos ^2(-x)$ $=\left(1-x^2\right)(-\sin x)\left(\cos ^2 x\right)$ $=-\left(1-x^2\right)(\sin x)\left(\cos ^2 x\right)$ So, $\quad f(x)=-f(-x)$ So, $f(x)$ is an odd function. Hence, $\int_{-\pi}^\pi\left(1-x^2\right) \sin x \cdot \cos ^2 x d x=0$

Hence, the correct answer is option: 'D'.

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