Math 106a Solutions
Quiz 3
10/8/10
Recall the integration by parts formula: Zu dv =uv โZv du.
1. Zexcos x dx
To start, apply integration by parts with
u= cos x=โdu =โsin x dx
dv =exdx =โv=ex
Zexcos x dx =excos xโZex(โsin x)dx
=excos x+Zexsin x dx (Now apply IBP again, with u= sin x&dv =exdx.)
=excos x+exsin xโZexcos x dx (Next, add Zexcos x dx to both sides.)
This gives
2Zexcos x dx =excos x+exsin x
Zexcos x dx =1
2(excos x+exsin x) + C
2. Z1
0
arctan x dx
Apply integration by parts with
u= arctan x=โdu =1
1 + x2dx
dv =dx =โv=x
Z1
0
arctan x dx =hxarctan xi1
0
โZ1
0
x
1 + x2dx (let w= 1 + x2, then 1
2dw =x dx)
=hxarctan xi1
0
โ1
2Z2
1
dw
w(since x=0=โw= 1 & x=1=โw= 2)
=hxarctan xi1
0
โ๎1
2ln |w|๎2
1
= (arctan 1 โ0) โ๎1
2ln 2 โ1
2ln 1๎=ฯ
4โln 2
2
1