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Answer: 0

Hint: Use $\frac{d}{d x} \sin x \text { and } \int x^{n} d x$

Given: $\int_{0}^{2 \pi} \cos ^{7} x \sin ^{4} x\; dx$

Solution:

\begin{aligned} &I=\int_{0}^{2 \pi} \cos ^{7} x \sin ^{4} x\; dx\\\\ &=\int_{0}^{2 \pi} \cos ^{6} x \sin ^{4} x \cdot \cos x\; dx \end{aligned}

\begin{aligned} &=\int_{0}^{2 \pi}\left(\cos ^{2} x\right)^{3} \sin ^{4} x \cdot \cos x\; dx \\\\ &=\int_{0}^{2 \pi}\left(1-\sin ^{2} x\right)^{3} \sin ^{4} x \cdot \cos x\; dx \end{aligned}

Put

\begin{aligned} &\sin x=t \\\\ &\cos x\; d x=d t \end{aligned}

\begin{aligned} &I=\int_{0}^{2 \pi}\left(1-t^{2}\right)^{3} t^{4} d t \\\\ &=\int_{0}^{2 \pi}\left(1-t^{6}-3 t^{2}+3 t^{4}\right) t^{4} d t \end{aligned}

\begin{aligned} &=\int_{0}^{2 \pi}\left(t^{4}-t^{10}-3 t^{6}+3 t^{8}\right) d t \\\\ &=\left[\frac{t^{5}}{5}-\frac{t^{11}}{11}-\frac{3 t^{7}}{7}+\frac{3 t^{9}}{9}\right]_{0}^{2 \pi} \end{aligned}

Where $t= sinx$

$=\left[\frac{\sin ^{5} x}{5}-\frac{\sin ^{11} x}{11}-\frac{3 \sin ^{7} x}{7}+\frac{3 \sin ^{9} x}{9}\right]_{0}^{2 \pi}$

We know  $\sin (2 \pi)=0$

\begin{aligned} =& \frac{0}{5}-\frac{0}{11}-\frac{0}{7}-\frac{0}{9} \\\\ =& 0 \end{aligned}

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