Answer Key for Quiz 2 (section A)
1. The simplest approach is to substitute w=x6before doing anything else. Then dw = 6x5dx, so
dw
6=x5dx and we have
(1) Zx11 sin ¡x6¢dx =Zx6sin ¡x6¢x5dx =Zwsin wdw
6=1
6Zwsin w dw.
Now integrate by parts with u=wand dv = sin w dw, so that du =dw and v=−cos wand we have
Zwsin w dw =w(−cos w)−Z(−cos w)dw
=−wcos w+Zcos w dw =−wcos w+ sin w+C.
Combining this with (1), we conclude that
Zx11 sin ¡x6¢dx =1
6Zwsin w dw =1
6(−wcos w+ sin w) + C=1
6£−x6cos ¡x6¢+ sin ¡x6¢¤+C.
The integral can be done without substituting in more or less the same way, but the substitution makes it
clearer what to do. The only way that a power of xtimes sin ¡x6¢will integrate nicely is if it integrates to a
multiple of cos ¡x6¢, so let’s first calculate the derivative of cos¡x6¢, which is −sin ¡x6¢6x5; in other words,
if we could take dv =−6x5sin ¡x6¢dx then vwould b e cos ¡x6¢. This suggests that we should rewrite
Zx11 sin ¡x6¢dx =Zx11
−6x5£−6x5sin ¡x6¢¤dx =Zµ−
1
6x6¶£−6x5sin ¡x6¢¤dx
and then take u=−
1
6x6, in which case du =−x5dx. Then we have
Zx11 sin ¡x6¢dx =µ−
1
6x6¶cos ¡x6¢−Zcos ¡x6¢¡−x5dx¢=−
1
6x6cos ¡x6¢+Zx5cos ¡x6¢dx.
Now we could substitute w=x6again, or stay with the same thought process as before: the only way this
integral could come out nicely is if it comes out to a multiple of sin¡x6¢, so we calculate the derivative of
sin ¡x6¢and get cos ¡x6¢6x5, which implies that
Zx11 sin ¡x6¢dx =−
1
6x6cos ¡x6¢+Zx5cos ¡x6¢dx =−
1
6x6cos ¡x6¢+1
6sin ¡x6¢+C
as before.
2. In both integrals one wants to get rid of the powers of xby repeatedly integrating by parts. This
means that Zx2e9xdx will require two integrations by parts, whereas Zx9e2xdx will take nine, so I hope
everybody picked the first one. If we take u=x2and dv =e9xdx, then du = 2xdx and v=e9x
9and we have
(1) Zx2e9xdx =x2µe9x
9¶−Zµe9x
9¶2x dx =x2e9x
9
−
2
9Zx e9xdx.
In the remaining integral we again take dv =e9xdx, so that u=x,du =dx and v=e9x
9and we have
Zx e9xdx =xµe9x
9¶−Zµe9x
9¶dx =x e9x
9
−
1
9Ze9xdx =x e9x
9
−
e9x
81 +C.
