Answer Key for Quiz 1 (section A)
1. If we substitute u= 8x−3 then du = 8 dx, so du
8=dx and we have
Zsin(8x−3) dx =Zsin udu
8=1
8Zsin u du =−1
8cos u+C=−1
8cos (8x−3) + C.
2. If we substitute u= 2x5+ 13 then du = 10x4dx, so du
10 =x4dx and the integral becomes
Z¡2x5+ 13¢43 x4dx =Zu43 du
10 =1
10 Zu43 du =1
10
u44
44 +C=1
440 ¡2x5+ 13¢44 +C.
3. Following the hint we have
Z1−sin x
cos xdx =Z1−sin x
cos x
1 + sin x
1 + sin xdx =Z1−sin2x
cos x
dx
1 + sin x=Zcos2x
cos x
dx
1 + sin x=Zcos x dx
1 + sin x.
Now substitute u= 1 + sin x, so that du = cos x dx and we have
Z1−sin x
cos xdx =Zcos x dx
1 + sin x=Zdu
u= ln |u|+C= ln |1 + sin x|+C.
For the extra credit we can say
ln |1 + sin x|+C=Z1−sin x
cos xdx =Zdx
cos x−Zsin x dx
cos x,
which implies that
Zdx
cos x=Zsin x dx
cos x+ ln |1 + sin x|+C.
But we can do the integral on the right side by letting v= cos x, in which case dv =−sin x dx or −dv =
sin x dx. Then we have
Zsin x dx
cos x=Z−dv
v=−ln |v|+C=−ln |cos x|+C.
Therefore Zdx
cos x= ln |1 + sin x| − ln |cos x|+C= ln ¯
¯
¯
¯
1 + sin x
cos x¯
¯
¯
¯
+C.
This is often written instead as Zsec x dx = ln |sec x+ tan x|+C,
where sec xdenotes the secant of x, which is the reciprocal of cos x. You would not learn from our textbook
that there is such a function as sec x, but some of you may have picked it up on the street somewhere.