Answer
$\frac{2}{3}{\sin ^3}\left( {\frac{t}{2}} \right) - \frac{2}{5}{\sin ^5}\left( {\frac{t}{2}} \right) + C$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int {{{\cos }^3}\left( {\frac{t}{2}} \right){{\sin }^2}\left( {\frac{t}{2}} \right)} dt \cr
& {\text{Split }}{\cos ^3}\left( {\frac{t}{2}} \right){\text{ as }}{\cos ^2}\left( {\frac{t}{2}} \right)\cos \left( {\frac{t}{2}} \right) \cr
&I= \int {{{\cos }^2}\left( {\frac{t}{2}} \right){{\sin }^2}\left( {\frac{t}{2}} \right)} \cos \left( {\frac{t}{2}} \right)dt \cr
& {\text{Rewrite }}{\cos ^2}\left( {\frac{t}{2}} \right){\text{ using }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& I = \int {\left[ {1 - {{\sin }^2}\left( {\frac{t}{2}} \right)} \right]{{\sin }^2}\left( {\frac{t}{2}} \right)} \cos \left( {\frac{t}{2}} \right)dt \cr
& = \int {\left[ {{{\sin }^2}\left( {\frac{t}{2}} \right) - {{\sin }^4}\left( {\frac{t}{2}} \right)} \right]} \cos \left( {\frac{t}{2}} \right)dt \cr
& \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \sin \left( {\frac{t}{2}} \right),{\text{ }}du = \frac{1}{2}\cos \left( {\frac{t}{2}} \right)dt,{\text{ 2}}du = \cos \left( {\frac{t}{2}} \right)dt \cr
& {\text{Substituting}} \cr
& I= \int {\left( {{u^2} - {u^4}} \right)} \left( 2 \right)du \cr
& = 2\int {\left( {{u^2} - {u^4}} \right)} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& I= 2\left( {\frac{{{u^3}}}{3} - \frac{{{u^5}}}{5}} \right) + C \cr
& = \frac{{2{u^3}}}{3} - \frac{{2{u^5}}}{5} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}\sin \left( {\frac{t}{2}} \right){\text{ for }}u \cr
& I= \frac{2}{3}{\sin ^3}\left( {\frac{t}{2}} \right) - \frac{2}{5}{\sin ^5}\left( {\frac{t}{2}} \right) + C \cr} $$
