Assignment #8

1)

A) ʃsin u du = -cos u + C
B) ʃcos u du = sin u +C
C) ʃtan u du = -ln |cos u| + C
D) ʃcot u du = ln |sin u| + C
E) ʃsec u du = ln |sec u + tan u| +C
F) ʃcsc u du = -ln |csc u + cot u| + C

2)

You would set "u" equal to "2x" because it is inside the function "u^1/2". du would equal 2dx but it cannot be achieved because of the (4x+1)dx.