Answer
$${\text{part a) }}\cos \left( {\pi - x} \right) + C{\text{ part b)}} - \frac{1}{{{x^5} + 1}} + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {\sin \left( {\pi - x} \right)} dx \cr
& {\text{substitute }}u = \pi - x,{\text{ }}du = - dx \cr
& \int {\sin \left( {\pi - x} \right)} dx = - \int {\sin u} du \cr
& {\text{find antiderivative }} \cr
& = - \left( { - \cos u} \right) + C \cr
& = \cos u + C \cr
& {\text{write in terms of }}x \cr
& = \cos \left( {\pi - x} \right) + C \cr
& \cr
& {\text{part b}} \cr
& \int {\frac{{5{x^4}}}{{{{\left( {{x^5} + 1} \right)}^2}}}} dx \cr
& {\text{substitute }}u = {x^5} + 1,{\text{ }}du = 5{x^4}dx \cr
& \int {\frac{{5{x^4}}}{{{{\left( {{x^5} + 1} \right)}^2}}}} dx = \int {\frac{{du}}{{{u^2}}}} \cr
& = \int {{u^{ - 2}}du} \cr
& {\text{find antiderivative }} \cr
& = \frac{{{u^{ - 1}}}}{{ - 1}} + C \cr
& = - \frac{1}{u} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{{{x^5} + 1}} + C \cr} $$
