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$\int_0^{2 \pi} \sin ^6 x \cos ^5 x d x$ is equal to
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Verified Answer
The correct answer is:
$0$
Let $\begin{aligned} I & =\int_0^{2 \pi} \sin ^6 x \cos ^5 x d x \\ I & =2 \int_0^\pi \sin ^6 x \cos ^5 x d x\end{aligned}$
$[\because f(2 \pi-x)=f(x)]$
Let $f(x)=\sin ^6 x \cos ^5 x$
$[\because f(\pi-x)=-f(x)]$
$\begin{aligned} f(\pi-x) & =\sin ^6(\pi-x) \cdot \cos ^5(\pi-x) \\ & =\sin ^6 x \cdot\left(-\cos ^5 x\right) \\ & =-\sin ^6 x \cdot \cos ^5 x \\ & =-f(x) \\ \therefore \quad I & =0\end{aligned}$
$[\because f(2 \pi-x)=f(x)]$
Let $f(x)=\sin ^6 x \cos ^5 x$
$[\because f(\pi-x)=-f(x)]$
$\begin{aligned} f(\pi-x) & =\sin ^6(\pi-x) \cdot \cos ^5(\pi-x) \\ & =\sin ^6 x \cdot\left(-\cos ^5 x\right) \\ & =-\sin ^6 x \cdot \cos ^5 x \\ & =-f(x) \\ \therefore \quad I & =0\end{aligned}$
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