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

Question

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

Aicharya · Accepted Answer

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

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

Solution