Q116 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

$$ \int_{-4 \pi}^{4 \pi} \tan ^9 x \sin ^6 x \cos ^3 x d x= $$

  • A
    $16 \times \frac{\pi}{2}$
  • B
    $8 \times \frac{2}{3}$
  • C
    $16 \times \frac{14}{17} \times \frac{12}{15} \times \ldots \times \frac{2}{3}$
  • Correct answer
    D
    0

Solution

$$ \begin{aligned} & \text {} \int_{-4 \pi}^{4 \pi} \tan ^9 x \sin ^6 x \cos ^3 x d x \\ & \text { here } f(x)=\tan ^9 x \sin ^6 x \cos ^3 x \\ & \Rightarrow f(-x)=[\tan (-x)]^9[\sin (-x)]^6[\cos (-x)]^3 \\ & \Rightarrow f(-x)=-\tan ^9 x \cdot \sin ^6 x \cos ^3 x=-f(x) \end{aligned} $$ $\mathrm{f}(\mathrm{x})$ is an odd function. $$ \therefore \int_{-4 \pi}^{4 \pi} \tan ^9 x \sin ^6 x \cos ^3 x d x=0 $$

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

Relevant Questions