Answer
$${\text{part a) }} - \frac{{{{\cot }^2}x}}{2} + C{\text{ part b)}}\frac{{{{\left( {1 + \sin t} \right)}^{10}}}}{{10}} + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {\cot x{{\csc }^2}} xdx \cr
& {\text{substitute }}u = \cot x,{\text{ }}du = - {\csc ^2}xdx \cr
& \int {\cot x{{\csc }^2}} xdx = \int u \left( { - du} \right) \cr
& = - \int u du \cr
& {\text{find antiderivative}} \cr
& = - \frac{{{u^2}}}{2} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{{{{\left( {\cot x} \right)}^2}}}{2} + C \cr
& = - \frac{{{{\cot }^2}x}}{2} + C \cr
& \cr
& {\text{part b}} \cr
& {\int {\left( {1 + \sin t} \right)} ^9}\cos tdt \cr
& {\text{substitute }}u = 1 + \sin t,{\text{ }}du = \cos tdt \cr
& {\int {\left( {1 + \sin t} \right)} ^9}\cos tdt = \int {{u^9}} du \cr
& {\text{find antiderivative}} \cr
& = \frac{{{u^{10}}}}{{10}} + C \cr
& {\text{write in terms of }}t \cr
& = \frac{{{{\left( {1 + \sin t} \right)}^{10}}}}{{10}} + C \cr} $$
