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Answer: $\frac{8}{3}$

Given:  $\int_{0}^{\pi} \sin ^{3} x(1+2 \operatorname{cox})(1+\cos x)^{2} d x$

Hint: Use trigonometric identity and then apply substitution method

Solution:  $\int_{0}^{\pi} \sin ^{3} x(1+2 \operatorname{cox})(1+\cos x)^{2} d x$

\begin{aligned} &\int_{0}^{\pi} \sin x\left(1-\cos ^{2} x\right)(1+2 \operatorname{cox})(1+\cos x)^{2} d x \\ & \end{aligned}

${\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \sin ^{2} x=1-\cos ^{2} x \end{array}\right]}$

Now,

Let

\begin{aligned} &\cos x=t \\ & \end{aligned}

$-\sin x d x=d t$          (Diff w.r.t to x)

\begin{aligned} &\Rightarrow-\int_{0}^{\pi}\left(1-t^{2}\right)(1+2 t)(1+t)^{2} d x \\ & \end{aligned}

$\Rightarrow-\int_{0}^{\pi}\left(1-t^{2}\right)(1+2 t)(1+t)^{2} d t \\$

$\Rightarrow-\int_{0}^{\pi}\left(1+t^{2}+2 t+2 t+2 t^{3}+4 t^{2}-2 t^{2}-t^{4}-2 t^{3}-2 t^{5}-4 t^{4}\right) d t$

\begin{aligned} &\Rightarrow-\int_{0}^{\pi}\left(1+4 t+4 t^{2}-2 t^{3}-5 t^{4}-2 t^{5}\right) d t \\ &\end{aligned}

$\Rightarrow-\left(t+2 t^{2}+\frac{4 t^{3}}{3}-\frac{t^{4}}{2}-t^{5}-\frac{t^{6}}{3}\right)_{-1}^{1} \\$

$\left\{\begin{array}{l} \therefore 0

\begin{aligned} &\Rightarrow\left(t+2 t^{2}+\frac{4 t^{3}}{3}-\frac{t^{4}}{2}-t^{5}-\frac{t^{6}}{3}\right)_{-1}^{1} \\ & \end{aligned}

$\Rightarrow\left(1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\right)_{-1}^{1}-\left((-1)+2-\frac{4}{3}-\frac{1}{2}+1-\frac{1}{3}\right)_{-1}^{1}$

\begin{aligned} &\Rightarrow 2-\frac{1}{2}+1-2+\frac{5}{3}+\frac{1}{2} \\ & \end{aligned}

$=1+\frac{5}{3}=\frac{8}{3}$

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