Answer
$${\text{part a) }}\frac{2}{7}{\left( {1 + x} \right)^{7/2}} - \frac{4}{5}{\left( {1 + x} \right)^{5/2}} + \frac{2}{3}{\left( {1 + x} \right)^{3/2}} + C{\text{ part b)}} - \cot \left( {\sin x} \right) + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {{x^2}\sqrt {1 + x} } dx \cr
& {\text{substitute }}u = x + 1,{\text{ }}du = dx \cr
& \int {{x^2}\sqrt {1 + x} } dx = \int {{{\left( {u - 1} \right)}^2}\sqrt u } du \cr
& {\text{expand and multiply}} \cr
& = \int {\left( {{u^2} - 2u + 1} \right){u^{1/2}}} du \cr
& = \int {\left( {{u^{5/2}} - 2{u^{3/2}} + {u^{1/2}}} \right)} du \cr
& {\text{find antiderivative by the power rule}} \cr
& \frac{{{u^{7/2}}}}{{7/2}} - 2\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) + \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& \frac{2}{7}{u^{7/2}} - \frac{4}{5}{u^{5/2}} + \frac{2}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x \cr
& \frac{2}{7}{\left( {1 + x} \right)^{7/2}} - \frac{4}{5}{\left( {1 + x} \right)^{5/2}} + \frac{2}{3}{\left( {1 + x} \right)^{3/2}} + C \cr
& \cr
& {\text{part b}} \cr
& \int {{{\left[ {\csc \left( {\sin x} \right)} \right]}^2}} \cos xdx \cr
& {\text{substitute }}u = \sin x,{\text{ }}du = \cos xdx \cr
& \int {{{\left[ {\csc \left( {\sin x} \right)} \right]}^2}} \cos xdx = \int {{{\csc }^2}udu} \cr
& {\text{use integration formulas from table 4}}{\text{.2}}{\text{.1}} \cr
& = - \cot u + C \cr
& {\text{write in terms of }}x \cr
& = - \cot \left( {\sin x} \right) + C \cr} $$
