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